The Equation of a Circle

A circle can be described geometrically as all points in a plane that are the same distance from a fixed point.

The fixed point is the center, and the fixed distance is the radius.

On a coordinate plane, this definition can be represented algebraically using the equation of a circle. The equation tells us exactly where the circle is located and how large it is.

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Standard Form of a Circle Equation

The standard form of the equation of a circle is:

(x − h)² + (y − k)² = r²

where:

• (h, k) is the center
• r is the radius
• r² is the square of the radius
• (x, y) represents any point on the circle

This equation is one of the most useful ways to describe a circle on a coordinate plane.

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Understanding the Equation

Consider:

(x − 3)² + (y − 2)² = 16

Compare this with:

(x − h)² + (y − k)² = r²

We can identify:

h = 3

k = 2

r² = 16

Therefore:

r = 4

The circle has:

center = (3, 2)

radius = 4

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Why the Circle Equation Works

The equation of a circle comes from the distance formula, which itself comes from the Pythagorean theorem.

Suppose:

• the center is (h, k)
• a point on the circle is (x, y)

The horizontal distance between the points is:

x − h

The vertical distance is:

y − k

Because the distance from the center to every point on the circle is r:

(x − h)² + (y − k)² = r²

This is exactly the standard equation of a circle.

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Recognizing a Circle Equation

An equation such as:

(x − 4)² + (y + 1)² = 25

is in standard circle form because it follows the pattern:

(x − h)² + (y − k)² = r²

Both variable expressions are squared and added together, and the right side represents the square of the radius.

From the equation we can determine the center and radius immediately.

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Reading the Center Carefully

One of the most important skills is reading the signs correctly.

Consider:

(x − 5)² + (y − 3)² = 49

The center is:

(5, 3)

Now consider:

(x + 5)² + (y − 3)² = 49

Rewrite mentally:

x + 5 = x − (−5)

Therefore, the center is:

(−5, 3)

The sign inside each bracket is the opposite of the corresponding center coordinate.

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The Sign Rule

If the equation contains:

(x − 4)²

then:

h = 4

If it contains:

(x + 4)²

then:

h = −4

Similarly:

(y − 7)² → k = 7

(y + 7)² → k = −7

A useful reminder is:

The signs inside the brackets are opposite the coordinates of the center.

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Worked Example: Identify Center and Radius

Consider:

(x − 2)² + (y + 6)² = 36

Compare with:

(x − h)² + (y − k)² = r²

For x:

h = 2

For y:

y + 6 = y − (−6)

so:

k = −6

For the radius:

r² = 36

r = √36

r = 6

Therefore:

center = (2, −6)

radius = 6

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The Radius Is Not r²

Consider:

(x + 1)² + (y − 4)² = 81

The number on the right is:

r² = 81

It is not the radius itself.

Take the square root:

r = √81

r = 9

Therefore:

center = (−1, 4)

radius = 9

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Worked Example with a Non-Perfect Square

Consider:

(x − 3)² + (y + 2)² = 20

The center is:

(3, −2)

and:

r² = 20

Therefore:

r = √20

Simplify:

r = √(4 × 5)

r = 2√5

An approximate decimal value is:

r ≈ 4.47

The radius does not have to be a whole number.

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Circles Centered at the Origin

The origin is:

(0, 0)

If a circle has its center at the origin, then:

h = 0

k = 0

The standard equation becomes:

x² + y² = r²

For example:

x² + y² = 25

has:

center = (0, 0)

radius = 5

This is an important special case.

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Writing an Equation from the Center and Radius

Suppose a circle has:

center = (4, 3)

radius = 5

Start with:

(x − h)² + (y − k)² = r²

Substitute:

h = 4

k = 3

r = 5

Therefore:

(x − 4)² + (y − 3)² = 25

Notice that the right side contains:

5² = 25

not simply 5.

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Worked Example with a Negative Coordinate

Write the equation of a circle with:

center = (−3, 2)

radius = 4

Start with:

(x − h)² + (y − k)² = r²

Substitute:

(x − (−3))² + (y − 2)² = 4²

Simplify:

(x + 3)² + (y − 2)² = 16

The negative x-coordinate produces a plus sign inside the bracket.

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Worked Example with Two Negative Coordinates

Write the equation for a circle with:

center = (−5, −2)

radius = 7

Substitute:

(x − (−5))² + (y − (−2))² = 7²

Simplify:

(x + 5)² + (y + 2)² = 49

This is why careful use of parentheses is helpful when the center contains negative coordinates.

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Worked Example Centered at the Origin

Suppose:

center = (0, 0)

radius = 8

Then:

(x − 0)² + (y − 0)² = 8²

Therefore:

x² + y² = 64

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Writing an Equation from the Diameter

Sometimes the diameter is given instead of the radius.

Suppose a circle has:

center = (2, 5)

diameter = 12

First find the radius:

r = d/2

r = 12/2

r = 6

Now:

r² = 36

Therefore:

(x − 2)² + (y − 5)² = 36

Always convert diameter to radius before writing the equation.

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Writing an Equation from a Center and a Point

Sometimes the radius is not given directly.

Suppose a circle has center:

(1, 2)

and passes through:

(4, 6)

The radius is the distance between these points.

Use the distance relationship:

r² = (4 − 1)² + (6 − 2)²

r² = 3² + 4²

r² = 9 + 16

r² = 25

Therefore:

r = 5

and the circle equation is:

(x − 1)² + (y − 2)² = 25

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Verifying Whether a Point Lies on a Circle

A point lies on a circle if its coordinates make the circle equation true.

Suppose:

(x − 2)² + (y − 1)² = 25

Does:

(5, 5)

lie on the circle?

Substitute:

x = 5

y = 5

(5 − 2)² + (5 − 1)² = 25

3² + 4² = 25

9 + 16 = 25

25 = 25

The equation is true.

Therefore:

(5, 5) lies on the circle.

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A Point That Does Not Lie on the Circle

Use the same circle:

(x − 2)² + (y − 1)² = 25

Test:

(4, 3)

Substitute:

(4 − 2)² + (3 − 1)²

= 2² + 2²

= 4 + 4

= 8

But:

8 ≠ 25

Therefore:

(4, 3) does not lie on the circle.

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Inside, On, or Outside the Circle

Substitution can tell us more than whether a point lies exactly on the circle.

For:

(x − h)² + (y − k)² = r²

calculate the left side for the point.

If:

left side = r²

the point is on the circle.

If:

left side < r²

the point is inside the circle.

If:

left side > r²

the point is outside the circle.

This works because the left side represents the square of the point's distance from the center.

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Worked Example: Inside, On, or Outside

Consider:

x² + y² = 25

Test:

(2, 3)

Calculate:

2² + 3² = 4 + 9 = 13

Since:

13 < 25

the point is:

inside the circle

Now test:

(3, 4)

3² + 4² = 9 + 16 = 25

Therefore:

(3, 4) is on the circle

Now test:

(5, 4)

5² + 4² = 25 + 16 = 41

Since:

41 > 25

the point is:

outside the circle

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Graphing a Circle from Its Equation

Suppose we want to graph:

(x − 2)² + (y − 3)² = 16

First identify:

center = (2, 3)

radius = 4

Plot the center:

(2, 3)

Then move 4 units:

• right
• left
• up
• down

from the center.

This gives four useful points on the circle.

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Finding the Four Main Points

For center:

(h, k)

and radius:

r

the four main points are:

Right: (h + r, k)

Left: (h − r, k)

Top: (h, k + r)

Bottom: (h, k − r)

For:

center = (2, 3)

radius = 4

we obtain:

Right: (6, 3)

Left: (−2, 3)

Top: (2, 7)

Bottom: (2, −1)

These points help us sketch the circle accurately.

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Graphing Example

Graph:

(x + 1)² + (y − 2)² = 9

Identify the center:

(−1, 2)

Find the radius:

r = √9

r = 3

Main points:

Right:

(−1 + 3, 2) = (2, 2)

Left:

(−1 − 3, 2) = (−4, 2)

Top:

(−1, 2 + 3) = (−1, 5)

Bottom:

(−1, 2 − 3) = (−1, −1)

Plot these points and draw a smooth circle through them.

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The Bounding Square

One way to improve a circle sketch is to imagine a square surrounding the circle.

For a circle with radius r:

• the circle extends r units left of the center
• r units right
• r units upward
• r units downward

The resulting square has:

side length = 2r

The circle fits exactly inside this square.

This helps prevent drawing a circle that is too wide, too narrow, or incorrectly centered.

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Finding the Circle's Extreme Coordinates

For a circle centered at:

(h, k)

with radius:

r

the smallest x-coordinate on the circle is:

h − r

The largest x-coordinate is:

h + r

The smallest y-coordinate is:

k − r

The largest y-coordinate is:

k + r

These values are useful when graphing and interpreting circles.

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Worked Example: Extreme Values

Consider:

(x − 4)² + (y + 2)² = 36

Center:

(4, −2)

Radius:

6

Leftmost x-value:

4 − 6 = −2

Rightmost x-value:

4 + 6 = 10

Lowest y-value:

−2 − 6 = −8

Highest y-value:

−2 + 6 = 4

This gives the overall position and size of the circle.

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How the Center Changes the Graph

Compare:

x² + y² = 16

and:

(x − 4)² + y² = 16

Both circles have:

r = 4

Therefore, they have the same size.

The first has center:

(0, 0)

The second has center:

(4, 0)

The second circle has been shifted:

4 units to the right

Changing h or k moves the circle without changing its size.

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Horizontal Shifts

Compare:

x² + y² = 25

with:

(x − 3)² + y² = 25

The second circle moves:

3 units right

Now consider:

(x + 3)² + y² = 25

The center is:

(−3, 0)

so the circle moves:

3 units left

Remember that the sign inside the bracket is opposite the direction of the center coordinate.

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Vertical Shifts

Consider:

x² + (y − 4)² = 9

The center is:

(0, 4)

so the circle has moved:

4 units upward

Now consider:

x² + (y + 4)² = 9

The center is:

(0, −4)

so the circle has moved:

4 units downward

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How the Radius Changes the Graph

Compare:

x² + y² = 4

x² + y² = 16

x² + y² = 36

Their radii are:

√4 = 2

√16 = 4

√36 = 6

Therefore, the circles become larger as r increases.

The center remains:

(0, 0)

Only the size changes.

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Same Center, Different Radii

Circles with the same center but different radii are called:

concentric circles

For example:

x² + y² = 4

x² + y² = 16

x² + y² = 36

all have center:

(0, 0)

but different radii.

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Writing an Equation from a Graph

Suppose a graph shows a circle with:

center = (3, −2)

The circle extends from the center to:

(8, −2)

The radius is the distance between these points:

r = 8 − 3

r = 5

Now use:

(x − h)² + (y − k)² = r²

Substitute:

(x − 3)² + (y + 2)² = 25

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Finding Radius from Two Graph Points

Suppose the center is:

(−2, 4)

and the rightmost point is:

(3, 4)

The horizontal distance is:

3 − (−2) = 5

Therefore:

r = 5

The equation is:

(x + 2)² + (y − 4)² = 25

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Connecting the Equation to Circle Properties

The equation tells us several geometric properties immediately.

For:

(x − 6)² + (y + 3)² = 49

we know:

center = (6, −3)

radius = 7

diameter = 14

leftmost point = (−1, −3)

rightmost point = (13, −3)

top point = (6, 4)

bottom point = (6, −10)

One equation contains a large amount of geometric information.

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Finding the Diameter from an Equation

Consider:

(x + 2)² + (y − 5)² = 100

First:

r² = 100

Therefore:

r = 10

The diameter is:

d = 2r

d = 20

So:

center = (−2, 5)

radius = 10

diameter = 20

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Verifying a Point Using Distance Instead

There are two equivalent ways to test a point.

For a circle centered at:

(2, 1)

with radius:

5

and the point:

(5, 5)

we can substitute into:

(x − 2)² + (y − 1)² = 25

or calculate the distance from the center:

distance² = (5 − 2)² + (5 − 1)²

distance² = 9 + 16

distance² = 25

Therefore:

distance = 5

The point lies exactly one radius from the center, so it lies on the circle.

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Why Squared Distance Is Often Enough

When checking whether a point lies on a circle, you usually do not need to calculate the square root.

For example, if:

r = 7

then:

r² = 49

If the squared distance from the point to the center is also:

49

the point lies on the circle.

This makes calculations faster.

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A Complete Example

Consider:

(x + 3)² + (y − 1)² = 64

Identify the center.

Because:

x + 3 = x − (−3)

the x-coordinate is:

−3

The y-coordinate is:

1

Therefore:

center = (−3, 1)

Now find the radius:

r² = 64

r = 8

Therefore:

radius = 8

and:

diameter = 16

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Graphing the Complete Example

For:

center = (−3, 1)

radius = 8

find the four main points.

Right:

(−3 + 8, 1)

(5, 1)

Left:

(−3 − 8, 1)

(−11, 1)

Top:

(−3, 1 + 8)

(−3, 9)

Bottom:

(−3, 1 − 8)

(−3, −7)

Plot the center and these four points, then draw a smooth circle.

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Complete Point Verification

Using:

(x + 3)² + (y − 1)² = 64

test the point:

(1, 7)

Substitute:

(1 + 3)² + (7 − 1)²

= 4² + 6²

= 16 + 36

= 52

Since:

52 < 64

the point is:

inside the circle

It is not on the circumference.

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Real-World Modeling

Circle equations can model circular objects and regions on coordinate systems.

Examples include:

• circular parks
• fountains
• radar ranges
• Ferris wheels
• circular tracks
• rotating machinery
• restricted zones
• communication ranges
• circular architectural designs

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Real-World Example: Circular Park

Suppose a circular park is represented on a coordinate map.

Its center is:

(20, 30)

and its radius is:

10 units

Its equation is:

(x − 20)² + (y − 30)² = 100

A location can be tested by substituting its coordinates into the equation.

If the left side equals 100, the location is:

on the park boundary

If it is less than 100:

inside the park

If it is greater than 100:

outside the park

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Real-World Example: Communication Range

Suppose a transmitter is located at:

(5, −2)

and has an idealized circular range of:

12 km

The boundary of its range can be modeled by:

(x − 5)² + (y + 2)² = 144

Locations satisfying:

(x − 5)² + (y + 2)² < 144

are inside the idealized coverage region.

The circle equation describes the boundary itself.

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Circle vs Filled Circular Region

This distinction is important.

The equation:

(x − h)² + (y − k)² = r²

describes only the:

circle boundary

The inequality:

(x − h)² + (y − k)² < r²

describes points:

inside the circle

The inequality:

(x − h)² + (y − k)² ≤ r²

describes:

the interior and the boundary

So an equation and an inequality represent different geometric sets.

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A Reliable Strategy for Reading a Circle Equation

Given:

(x − h)² + (y − k)² = r²

use these steps:

1. Check that the equation is in standard form.
2. Identify the x-expression.
3. Reverse its sign to find h.
4. Identify the y-expression.
5. Reverse its sign to find k.
6. Write the center as (h, k).
7. Identify r².
8. Take the positive square root to find r.
9. Check that the radius is positive.
10. Use the center and radius to interpret or graph the circle.

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A Reliable Strategy for Writing a Circle Equation

When given a center and radius:

1. Write the standard form:

   (x − h)² + (y − k)² = r²

2. Identify h and k.
3. Substitute the center coordinates.
4. Be careful with negative coordinates.
5. Substitute the radius.
6. Square the radius.
7. Simplify the signs.
8. Check that the equation gives the correct center and radius.

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A Reliable Strategy for Checking a Point

To determine whether (a, b) lies on a circle:

1. Replace x with a.
2. Replace y with b.
3. Evaluate both squared expressions.
4. Add the results.
5. Compare with r².

If:

result = r² → on

result < r² → inside

result > r² → outside

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A Reliable Strategy for Graphing a Circle

To graph a circle:

1. Identify the center (h, k).
2. Plot the center.
3. Find the radius r.
4. Move r units right from the center.
5. Move r units left.
6. Move r units up.
7. Move r units down.
8. Plot these four points.
9. Sketch a smooth circle through them.
10. Check that the circle is symmetric about horizontal and vertical lines through its center.

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Common Mistakes

Using the signs inside the brackets directly

In:

(x + 4)² + (y − 2)² = 9

the center is:

(−4, 2)

not (4, −2).

Using r² as the radius

If:

r² = 49

then:

r = 7

not 49.

Forgetting to square the radius

If r = 6, the equation must have:

r² = 36

on the right.

Forgetting parentheses with negative coordinates

If h = −3:

x − (−3)

becomes:

x + 3

Thinking x² + y² = 25 has center (25, 25)

The missing h and k values are both zero.

Therefore, the center is:

(0, 0)

Graphing the radius as the diameter

If r = 4, move 4 units from the center, not 8.

Testing only one coordinate

Both x and y coordinates must be substituted.

Taking a negative radius

Radius represents a distance and is positive.

If r² = 36:

r = 6

for the geometric radius.

Assuming every point inside the circle lies on the circle

Only points exactly r units from the center lie on the circumference.

Drawing the circle from appearance alone

Use the center and radius to establish accurate reference points first.

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Key Terms

Circle — set of all points in a plane located the same distance from a fixed center.

Center — fixed point (h, k) from which every point on the circle is the same distance.

Radius — distance from the center to any point on the circle.

Diameter — distance across the circle through its center; equal to 2r.

Standard form of a circle — (x − h)² + (y − k)² = r².

h — x-coordinate of the center.

k — y-coordinate of the center.

r — radius of the circle.

r² — square of the radius and the value appearing on the right side of the standard equation.

Origin — coordinate point (0, 0).

Coordinate plane — two-dimensional system used to locate points using x- and y-coordinates.

Distance formula — relationship used to calculate the distance between two coordinate points.

Concentric circles — circles sharing the same center but having different radii.

Boundary — set of points exactly on the circumference.

Interior — points whose distance from the center is less than the radius.

Exterior — points whose distance from the center is greater than the radius.

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Key Takeaways

• The standard equation of a circle is (x − h)² + (y − k)² = r².
• The center is (h, k).
• The signs inside the brackets appear opposite to the signs of the center coordinates.
• The number on the right side is r², not r.
• Take the positive square root of r² to find the radius.
• A circle centered at the origin has equation x² + y² = r².
• Given the center and radius, substitute h, k, and r into the standard form.
• Negative center coordinates create addition signs inside the corresponding brackets.
• If the diameter is given, divide by 2 before writing the equation.
• If a center and a point on the circle are given, the radius can be found using the distance between them.
• A point lies on a circle when its coordinates make the circle equation true.
• If the squared distance from the center is less than r², the point is inside.
• If the squared distance is greater than r², the point is outside.
• To graph a circle, first plot its center and then locate points one radius away in the four main directions.
• Changing h moves a circle horizontally.
• Changing k moves a circle vertically.
• Changing r changes the size of the circle.
• Circles with the same center but different radii are concentric.
• The circle equation describes the boundary, while an inequality can describe the interior.
• Circle equations connect algebra, coordinate geometry, distance, and real-world spatial modeling.

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Check Your Understanding

1. State the standard form of the equation of a circle and explain what h, k, and r represent.

2. Identify the center and radius of:

   (x − 3)² + (y − 5)² = 49

3. Identify the center and radius of:

   (x + 4)² + (y − 2)² = 81

4. Identify the center and radius of:

   x² + y² = 100

5. Write the equation of a circle with center (2, 6) and radius 5.
6. Write the equation of a circle with center (−3, 4) and radius 7.
7. Write the equation of a circle centered at (−2, −5) with diameter 12.
8. A circle has center (1, 2) and passes through (4, 6). Determine its radius and write its equation.
9. Determine whether (3, 4) lies on the circle x² + y² = 25.
10. Determine whether (2, 1) is inside, on, or outside the circle x² + y² = 16.
11. For (x − 5)² + (y + 2)² = 36, determine the center, radius, diameter, leftmost point, rightmost point, highest point, and lowest point.
12. Describe how the graph of x² + y² = 25 changes when the equation becomes (x − 4)² + y² = 25.
13. Describe how the graph of (x − 2)² + (y + 3)² = 16 changes if the right side becomes 64.
14. A student says that (x + 6)² + (y − 1)² = 25 has center (6, −1) and radius 25. Identify both errors and give the correct center and radius.
15. A circular safety zone has center (10, −4) and radius 8 km. Write an equation for its boundary and explain how you could determine whether a location at (14, 2) lies inside, on, or outside the safety zone.

Finding Centers and Radii

The equation of a circle contains two of its most important geometric properties:

• its center, which determines its location
• its radius, which determines its size

When a circle is written in standard form, both pieces of information can usually be identified directly.

The standard form is:

(x − h)² + (y − k)² = r²

where:

• (h, k) = center
• r = radius

Learning to read this equation quickly allows us to graph, construct, compare, and analyze circles.

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Identifying the Center

In:

(x − h)² + (y − k)² = r²

the center is:

(h, k)

For example:

(x − 4)² + (y − 3)² = 25

gives:

h = 4

k = 3

Therefore:

center = (4, 3)

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Watch the Signs

The signs inside the brackets are opposite the coordinates of the center.

For example:

(x + 5)² + (y − 2)² = 16

can be thought of as:

[x − (−5)]² + (y − 2)² = 16

Therefore:

center = (−5, 2)

This sign reversal is one of the most important ideas when reading circle equations.

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Reading Different Sign Combinations

Consider the following examples:

The number inside each bracket tells you the coordinate after you reverse its sign.

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Finding the Radius

The right side of the standard equation is:

r²

Therefore, to find the radius, take the positive square root.

For example:

(x − 2)² + (y + 1)² = 49

gives:

r² = 49

r = √49

r = 7

The radius is not 49.

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Why the Radius Is Positive

Solving:

r² = 49

algebraically could produce:

r = ±7

However, the geometric radius represents a:

distance

A distance cannot be negative.

Therefore, the radius of the circle is:

r = 7

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Worked Example: Center and Radius

Find the center and radius of:

(x − 6)² + (y + 2)² = 81

Identify the center:

x − 6 → h = 6

y + 2 → k = −2

Therefore:

center = (6, −2)

Now find the radius:

r² = 81

r = √81

radius = 9

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Worked Example with Two Negative Coordinates

Find the center and radius of:

(x + 4)² + (y + 7)² = 36

The center is:

(−4, −7)

because:

x + 4 = x − (−4)

y + 7 = y − (−7)

The radius is:

r = √36

r = 6

Therefore:

center = (−4, −7)

radius = 6

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Worked Example with a Non-Perfect Square

Consider:

(x − 1)² + (y − 5)² = 40

Center:

(1, 5)

Radius:

r = √40

Simplify:

r = √(4 × 10)

r = 2√10

Approximately:

r ≈ 6.32

Not every circle has a whole-number radius.

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Circles Centered at the Origin

If the center is:

(0, 0)

then:

h = 0

k = 0

and the equation simplifies to:

x² + y² = r²

For example:

x² + y² = 64

has:

center = (0, 0)

radius = 8

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When Only One Coordinate Is Zero

Consider:

x² + (y − 5)² = 16

The x-part can be written:

(x − 0)²

Therefore:

h = 0

k = 5

The center is:

(0, 5)

and:

r = 4

Now consider:

(x + 3)² + y² = 100

This gives:

center = (−3, 0)

radius = 10

A missing number inside a squared variable term means the corresponding center coordinate is zero.

---

Connecting the Equation to the Graph

Once the center and radius are known, the circle can be located on a coordinate plane.

For:

(x − 3)² + (y + 2)² = 25

we have:

center = (3, −2)

radius = 5

From the center, the circle extends 5 units:

• left
• right
• up
• down

This provides four useful points for constructing the circle.

---

Finding the Four Main Points

For a circle with center:

(h, k)

and radius:

r

the four main points are:

Right:

(h + r, k)

Left:

(h − r, k)

Top:

(h, k + r)

Bottom:

(h, k − r)

These points help determine the circle's location and size.

---

Worked Example: Constructing a Circle

Consider:

(x − 2)² + (y − 4)² = 9

Center:

(2, 4)

Radius:

3

Find the four main points.

Right:

(2 + 3, 4)

(5, 4)

Left:

(2 − 3, 4)

(−1, 4)

Top:

(2, 4 + 3)

(2, 7)

Bottom:

(2, 4 − 3)

(2, 1)

Plot these four points around the center and draw a smooth circle through them.

---

Constructing a Circle from Coordinate Information

Suppose you are told:

center = (−3, 2)

and:

radius = 4

You can construct the circle by first plotting:

(−3, 2)

Then move 4 units in each main direction.

Right:

(1, 2)

Left:

(−7, 2)

Top:

(−3, 6)

Bottom:

(−3, −2)

The equation is:

(x + 3)² + (y − 2)² = 16

---

Constructing a Circle from Center and Diameter

Suppose a circle has:

center = (5, −1)

diameter = 12

First calculate the radius:

r = d/2

r = 12/2

r = 6

Therefore:

r² = 36

The equation is:

(x − 5)² + (y + 1)² = 36

This circle extends 6 coordinate units in every direction from its center.

---

Constructing a Circle from Center and One Point

Suppose the center is:

(2, 3)

and the circle passes through:

(6, 6)

The radius is the distance from the center to the point.

Instead of immediately finding r, we can first find r²:

r² = (6 − 2)² + (6 − 3)²

r² = 4² + 3²

r² = 16 + 9

r² = 25

Therefore:

r = 5

The circle equation is:

(x − 2)² + (y − 3)² = 25

---

Why Coordinate Information Is Enough

A circle is completely determined if we know:

• its center
• its radius

Once those two properties are known, there is only one possible circle.

Similarly, if we know the center and one point on the circle, the distance between those points determines the radius.

This is why coordinate information can be used to construct an exact circle.

---

Finding the Diameter from an Equation

Consider:

(x + 2)² + (y − 8)² = 121

Center:

(−2, 8)

Radius:

r = √121

r = 11

Diameter:

d = 2r

d = 22

Therefore:

diameter = 22

---

Finding the Circle's Horizontal Range

Consider:

(x − 4)² + (y − 1)² = 36

Center:

(4, 1)

Radius:

6

The circle extends from:

h − r

to:

h + r

Therefore:

minimum x = 4 − 6 = −2

maximum x = 4 + 6 = 10

So the circle extends horizontally from:

x = −2 to x = 10

---

Finding the Circle's Vertical Range

Using the same circle:

center = (4, 1)

radius = 6

The minimum y-coordinate is:

k − r

1 − 6 = −5

The maximum y-coordinate is:

k + r

1 + 6 = 7

Therefore, the circle extends vertically from:

y = −5 to y = 7

---

Finding Extreme Points

For:

(x − 4)² + (y − 1)² = 36

the extreme points are:

Left:

(−2, 1)

Right:

(10, 1)

Top:

(4, 7)

Bottom:

(4, −5)

These points are useful when solving problems involving the position of a circle.

---

Comparing Circle Locations

Consider:

Circle A:

(x − 2)² + (y − 3)² = 16

Circle B:

(x − 7)² + (y − 3)² = 16

Circle A has center:

(2, 3)

Circle B has center:

(7, 3)

Both have:

r = 4

Therefore:

• they have the same size
• Circle B is 5 units to the right of Circle A

Only the center has changed.

---

Comparing Circle Sizes

Consider:

Circle A:

(x − 2)² + (y − 3)² = 16

Circle B:

(x − 2)² + (y − 3)² = 49

Both have center:

(2, 3)

Circle A:

r = 4

Circle B:

r = 7

Therefore:

Circle B is larger

The circles are concentric because they share the same center.

---

Comparing Both Location and Size

Consider:

Circle A:

(x + 1)² + (y − 4)² = 25

Circle B:

(x − 5)² + (y + 2)² = 64

Circle A:

center = (−1, 4)

radius = 5

Circle B:

center = (5, −2)

radius = 8

Therefore:

• the centers are different
• Circle B is larger
• Circle B is positioned farther right and lower than Circle A

Comparing circle equations lets us compare both:

location and size

---

Comparing Circles Efficiently

When comparing two circle equations, create a simple summary.

Then ask:

• Do they have the same center?
• Do they have the same radius?
• Which is larger?
• How are their centers positioned relative to each other?

---

Congruent Circles

Two circles are congruent if they have the same radius.

They do not need to have the same center.

For example:

(x − 2)² + y² = 25

and:

(x + 7)² + (y − 4)² = 25

both have:

r = 5

Therefore, the circles are:

congruent

They have the same size but different locations.

---

Concentric Circles

Two or more circles are concentric if they have the same center but different radii.

For example:

(x − 3)² + (y + 1)² = 9

and:

(x − 3)² + (y + 1)² = 64

both have center:

(3, −1)

Their radii are:

3 and 8

Therefore, they are:

concentric circles

---

Same Center and Same Radius

Suppose:

Circle A:

(x − 4)² + (y + 2)² = 36

Circle B:

(x − 4)² + (y + 2)² = 36

Both have:

center = (4, −2)

radius = 6

They are not two different circles.

They are:

the same circle

A circle is uniquely determined by its center and radius.

---

Distance Between Circle Centers

Sometimes comparing circle locations requires finding the distance between their centers.

Suppose:

Circle A has center:

(1, 2)

Circle B has center:

(5, 5)

The horizontal change is:

5 − 1 = 4

The vertical change is:

5 − 2 = 3

Using the distance formula:

distance = √(4² + 3²)

distance = √25

distance = 5

The centers are 5 units apart.

---

Comparing Distance with Radii

Knowing the distance between centers can help us reason about whether circles are separate, touching, or overlapping.

Suppose two circles have:

r₁ = 3

r₂ = 2

and the centers are:

5 units apart.

Then:

r₁ + r₂ = 3 + 2

r₁ + r₂ = 5

Since the center distance equals the sum of the radii, the circles touch externally at:

exactly one point

---

Separate Circles

If the distance between centers is greater than the sum of the radii:

d > r₁ + r₂

the circles are separate.

For example:

r₁ = 3

r₂ = 4

center distance = 10

Since:

10 > 3 + 4

10 > 7

the circles do not touch or overlap.

---

Intersecting Circles

If:

|r₁ − r₂| < d < r₁ + r₂

the circles intersect at two points.

For example:

r₁ = 5

r₂ = 4

d = 6

Difference:

|5 − 4| = 1

Sum:

5 + 4 = 9

Since:

1 < 6 < 9

the circles intersect at:

two points

---

External Tangency

Two circles are externally tangent when they touch at exactly one point from the outside.

This occurs when:

d = r₁ + r₂

Example:

r₁ = 5

r₂ = 3

d = 8

Since:

5 + 3 = 8

the circles are externally tangent.

---

Internal Tangency

One circle may lie inside another and touch it at exactly one point.

This occurs when:

d = |r₁ − r₂|

For example:

r₁ = 8

r₂ = 3

d = 5

Since:

|8 − 3| = 5

the circles are:

internally tangent

---

One Circle Inside Another

If:

d < |r₁ − r₂|

the smaller circle lies completely inside the larger circle without touching it.

For example:

r₁ = 8

r₂ = 3

d = 2

Since:

2 < |8 − 3|

2 < 5

the smaller circle lies completely inside the larger circle.

---

Circle Position Summary

Let:

d = distance between centers

Then:

This provides a powerful way to solve problems involving the location and size of circles.

---

Worked Comparison Problem

Consider:

Circle A:

x² + y² = 16

Circle B:

(x − 7)² + y² = 9

Circle A:

center = (0, 0)

radius = 4

Circle B:

center = (7, 0)

radius = 3

Distance between centers:

d = 7

Sum of radii:

r₁ + r₂ = 4 + 3 = 7

Therefore:

d = r₁ + r₂

The circles are:

externally tangent

---

Worked Overlap Problem

Circle A:

(x − 1)² + (y − 2)² = 25

Circle B:

(x − 5)² + (y − 2)² = 16

Centers:

A = (1, 2)

B = (5, 2)

Distance between centers:

d = 4

Radii:

r₁ = 5

r₂ = 4

Calculate:

|5 − 4| = 1

5 + 4 = 9

Since:

1 < 4 < 9

the circles:

intersect at two points

---

Real-World Example: Communication Zones

Suppose two wireless transmitters are modeled using circular coverage areas.

Transmitter A:

center = (2, 4)

radius = 6 km

Transmitter B:

center = (10, 4)

radius = 5 km

The centers are:

8 km apart.

The sum of the radii is:

6 + 5 = 11 km

The difference is:

|6 − 5| = 1 km

Since:

1 < 8 < 11

the two circular coverage boundaries intersect.

Their coverage regions also overlap.

This type of model can help analyze service areas.

---

Real-World Example: Circular Safety Zones

Suppose a safety zone has equation:

(x − 20)² + (y − 15)² = 100

The center is:

(20, 15)

The radius is:

10 units

Therefore, the zone's boundary is 10 units from the central location in every direction.

Its diameter is:

20 units

A point can be tested to determine whether it lies inside, on, or outside the safety boundary.

---

Testing a Location

Using:

(x − 20)² + (y − 15)² = 100

consider the location:

(26, 23)

Calculate:

(26 − 20)² + (23 − 15)²

= 6² + 8²

= 36 + 64

= 100

Therefore:

the point lies exactly on the circle

It is exactly 10 units from the center.

---

Inside or Outside a Circle

For:

(x − h)² + (y − k)² = r²

a point is:

inside if:

(x − h)² + (y − k)² < r²

on if:

(x − h)² + (y − k)² = r²

outside if:

(x − h)² + (y − k)² > r²

This allows us to solve many location problems without graphing.

---

Solving a Boundary Problem

A circular park has center:

(5, 6)

and radius:

8 m

A lamp is located at:

(9, 9)

Determine whether the lamp is inside the park boundary.

Calculate its squared distance from the center:

(9 − 5)² + (9 − 6)²

= 4² + 3²

= 16 + 9

= 25

Compare with:

r² = 8² = 64

Since:

25 < 64

the lamp is:

inside the circular park

---

Solving a Location Problem Without Square Roots

Notice that we did not calculate:

√25

We compared:

distance² = 25

with:

radius² = 64

Because:

25 < 64

we immediately know that the point is inside.

Comparing squared distances often makes circle problems faster and simpler.

---

Translating a Circle

Suppose:

x² + y² = 25

has center:

(0, 0)

and radius:

5.  

Now consider:

(x − 4)² + (y + 3)² = 25

The new center is:

(4, −3)

The radius remains:

5

Therefore, the circle has been moved:

• 4 units right
• 3 units down

without changing its size.

---

Changing Only the Radius

Compare:

(x − 2)² + (y − 1)² = 9

and:

(x − 2)² + (y − 1)² = 49

Both centers are:

(2, 1)

The first radius is:

3

The second radius is:

7

Therefore, the second circle is larger but has not moved.

---

Changing Only the Center

Compare:

(x − 2)² + (y − 1)² = 25

and:

(x + 4)² + (y − 6)² = 25

Both radii are:

5

Therefore, the circles are congruent.

However, their centers are:

(2, 1)

and:

(−4, 6)

so their locations are different.

---

A Reliable Strategy for Finding Center and Radius

When given a standard circle equation:

1. Compare it with (x − h)² + (y − k)² = r².
2. Identify the x-expression.
3. Reverse the sign to find h.
4. Identify the y-expression.
5. Reverse the sign to find k.
6. Write the center as (h, k).
7. Identify r².
8. Take the positive square root.
9. State the radius.
10. Check whether the result makes sense.

---

A Reliable Strategy for Constructing a Circle

When given coordinate information:

1. Identify or calculate the center.
2. Identify or calculate the radius.
3. Plot the center.
4. Move r units right.
5. Move r units left.
6. Move r units upward.
7. Move r units downward.
8. Plot the four extreme points.
9. Draw a smooth circle.
10. Write the equation if required.

---

A Reliable Strategy for Comparing Circles

For each circle:

1. identify the center
2. identify the radius
3. compare the center coordinates
4. compare the radii
5. determine whether they are the same size
6. determine how their locations differ
7. calculate the distance between centers if necessary
8. compare that distance with the radii
9. determine whether the circles are separate, tangent, intersecting, or nested

---

Common Mistakes

Reading the signs directly

For:

(x + 4)² + (y − 3)² = 25

the center is:

(−4, 3)

not (4, −3).

Using r² as the radius

If:

r² = 64

then:

r = 8

not 64.

Using a negative radius

Radius is a distance and must be positive.

Forgetting that x² means (x − 0)²

If no horizontal shift appears, the x-coordinate of the center is zero.

Forgetting that y² means (y − 0)²

If no vertical shift appears, the y-coordinate of the center is zero.

Using the diameter as the radius

If d = 14:

r = 7

Moving the circle when only the radius changes

Changing r changes size, not center.

Changing the size when only the center changes

If r² remains unchanged, the circle remains the same size.

Comparing the right-side numbers instead of the radii

For circle size, compare:

√r²

although larger positive r² also corresponds to a larger radius.

Judging circle relationships only from a sketch

Coordinate calculations are more reliable than appearance.

---

Key Terms

Center — fixed point (h, k) that determines the location of a circle.

Radius — distance from the center to every point on the circle.

Diameter — twice the radius.

Standard form — circle equation written as (x − h)² + (y − k)² = r².

Origin — point (0, 0).

Coordinate plane — system for locating points using x- and y-coordinates.

Extreme points — leftmost, rightmost, highest, and lowest points of a circle.

Congruent circles — circles with equal radii.

Concentric circles — circles sharing the same center.

Distance between centers — straight-line distance connecting the centers of two circles.

External tangency — two circles touching externally at exactly one point.

Internal tangency — one circle inside another while touching at exactly one point.

Intersecting circles — circles whose circumferences share two points.

Squared distance — distance calculation before taking the square root; useful for comparing a point with r².

Translation — movement of a figure without changing its size or shape.

---

Key Takeaways

• A circle in standard form is written as (x − h)² + (y − k)² = r².
• The center is (h, k).
• Signs inside the brackets are opposite the corresponding center coordinates.
• The right side represents r², not r.
• Radius is found by taking the positive square root of the right side.
• A circle centered at the origin has equation x² + y² = r².
• Missing horizontal or vertical shifts correspond to center coordinates of zero.
• A circle is completely determined by its center and radius.
• If the center and one point on the circle are known, the radius can be calculated using distance.
• The radius determines the size of a circle.
• The center determines its location.
• Circles with equal radii are congruent.
• Circles with the same center are concentric if their radii differ.
• Circles with the same center and same radius are the same circle.
• The four extreme points can be found by adding and subtracting the radius from the center coordinates.
• Comparing the distance between centers with the radii can determine whether circles are separate, tangent, intersecting, or nested.
• A point's squared distance from the center can determine whether it lies inside, on, or outside a circle.
• Circle equations can model real-world boundaries, ranges, zones, and circular structures.
• Algebraic information in a circle equation provides precise information about both location and size.

---

Check Your Understanding

1. Identify the center and radius of (x − 4)² + (y − 7)² = 36.
2. Identify the center and radius of (x + 3)² + (y − 5)² = 81.
3. Identify the center and radius of (x + 6)² + (y + 2)² = 49.
4. Determine the center, radius, and diameter of x² + y² = 144.
5. Determine the center and radius of x² + (y − 8)² = 25.
6. Write the equation of a circle with center (−4, 3) and radius 6.
7. A circle has center (5, −2) and diameter 18. Write its equation.
8. A circle has center (1, 2) and passes through (4, 6). Determine its radius and equation.
9. For (x − 3)² + (y + 4)² = 16, determine the leftmost, rightmost, highest, and lowest points.
10. Compare (x − 2)² + (y − 1)² = 25 and (x + 5)² + (y − 1)² = 25. How are the circles alike, and how are they different?
11. Compare (x − 4)² + (y + 2)² = 9 and (x − 4)² + (y + 2)² = 64. Describe their centers, radii, and relationship.
12. Two circles have radii 5 and 3, and their centers are 8 units apart. Describe the relationship between the circles and justify your answer.
13. Two circles have radii 7 and 4, and their centers are 5 units apart. Determine whether they are separate, tangent, intersecting, or nested.
14. A circular safety boundary has center (10, 6) and radius 8 units. Determine whether the point (16, 10) is inside, on, or outside the boundary.
15. A student says that (x + 5)² + (y − 2)² = 100 has center (5, −2) and radius 100. Identify both errors, give the correct center and radius, and explain how the equation determines the circle's location and size.

Applications of Circle Geometry

Circle geometry is used whenever objects, paths, boundaries, or motion can be modeled as circular. A circle equation can describe both the location and size of a circular object.

The standard equation is:

(x − h)² + (y − k)² = r²

where:

• (h, k) is the center
• r is the radius
• (x, y) represents points on the circular boundary

This allows circular situations to be studied using algebra, geometry, measurement, and coordinate systems.

---

Why Circle Geometry Is Useful

Circular shapes occur throughout science, engineering, transportation, architecture, and technology.

Examples include:

• wheels and gears
• Ferris wheels
• circular tracks
• radar ranges
• communication zones
• rotating machinery
• roundabouts
• satellites and idealized orbital paths
• circular buildings and structures
• pipes and tunnels
• sensors and scanning systems

Coordinate geometry allows these objects to be represented mathematically.

Once a circular situation has been modeled with an equation, we can calculate distances, test locations, compare objects, and make predictions.

---

Turning a Real Situation into a Mathematical Model

Suppose a circular fountain has its center at coordinate:

(8, 5)

and radius:

4 m

The boundary can be represented by:

(x − 8)² + (y − 5)² = 16

This equation tells us:

• where the fountain is located
• how large it is
• which points lie on its edge

The equation is a mathematical model of the physical fountain.

---

Choosing a Coordinate System

Before modeling a real object, we need a coordinate system.

We might choose:

• x = east-west position
• y = north-south position

or:

• x = horizontal distance
• y = vertical distance

We also need a scale.

For example:

1 coordinate unit = 1 metre

or:

1 coordinate unit = 10 km

Without a scale, the coordinates do not tell us the actual physical size of the object.

---

Modeling a Circular Object

Suppose a circular storage tank has center:

(12, 8)

and radius:

6 m

Its boundary can be modeled as:

(x − 12)² + (y − 8)² = 36

The tank extends:

6 m left

6 m right

6 m upward

6 m downward

from its center.

Therefore its extreme coordinates are:

Left: (6, 8)

Right: (18, 8)

Top: (12, 14)

Bottom: (12, 2)

---

Boundary vs Entire Circular Region

An important distinction must be made when modeling real objects.

The equation:

(x − h)² + (y − k)² = r²

represents the circular boundary.

If the entire region inside the circle is included, we use:

(x − h)² + (y − k)² ≤ r²

For example:

(x − 8)² + (y − 5)² ≤ 16

represents the fountain and all points inside its circular boundary.

---

Testing Whether a Location Is Inside a Circular Region

Suppose a circular safety zone has:

center = (4, 3)

radius = 5 km

Its boundary is:

(x − 4)² + (y − 3)² = 25

A location is at:

(7, 7)

Calculate:

(7 − 4)² + (7 − 3)²

= 3² + 4²

= 9 + 16

= 25

Since the result equals r²:

the location lies exactly on the boundary

---

Inside, On, or Outside

For a circle:

(x − h)² + (y − k)² = r²

calculate:

D² = (x − h)² + (y − k)²

Then compare D² with r².

If:

D² < r² → inside

D² = r² → on the boundary

D² > r² → outside

Using squared distances avoids unnecessary square-root calculations.

---

Worked Location Example

A sensor has an idealized detection radius of:

10 m

and is located at:

(6, 4)

A moving object is detected at:

(12, 10)

Calculate its squared distance from the sensor:

D² = (12 − 6)² + (10 − 4)²

D² = 6² + 6²

D² = 36 + 36

D² = 72

Sensor radius squared:

r² = 100

Since:

72 < 100

the object is:

inside the detection region

---

Why Squared Distance Is Useful in Technology

Computers often need to determine whether points are inside circular regions.

Instead of calculating:

D = √[(x − h)² + (y − k)²]

and then comparing D with r, software can compare:

D² with r²

This avoids calculating a square root.

For a single calculation this may not matter much, but systems performing thousands or millions of calculations can benefit from efficient mathematical methods.

---

Communication and Coverage Areas

Wireless transmitters are sometimes simplified as having circular coverage areas.

Suppose a transmitter is located at:

(15, 20)

with an idealized range of:

12 km

The coverage boundary can be modeled by:

(x − 15)² + (y − 20)² = 144

Locations satisfying:

(x − 15)² + (y − 20)² < 144

are inside the idealized coverage area.

---

Limitations of a Circular Coverage Model

Real wireless signals usually do not form perfect circles.

Signal strength can be affected by:

• buildings
• hills
• interference
• antenna design
• weather
• transmitter power

Therefore, a circular model is often an idealization.

A mathematical model does not need to reproduce reality perfectly to be useful. It needs to be accurate enough for its intended purpose.

---

Radar and Detection Systems

Radar and other detection systems often use distance from a known location.

Suppose a radar station is at:

(20, 30)

and can detect objects up to:

50 km

Its idealized maximum range boundary is:

(x − 20)² + (y − 30)² = 2500

because:

50² = 2500

An object's coordinates can be substituted into the expression to determine whether it is within range.

---

Comparing Two Coverage Areas

Suppose two transmitters have circular coverage regions.

Transmitter A:

center = (0, 0)

radius = 8 km

Transmitter B:

center = (12, 0)

radius = 6 km

The distance between centers is:

12 km

The sum of the radii is:

8 + 6 = 14 km

The difference is:

|8 − 6| = 2 km

Since:

2 < 12 < 14

the circular boundaries intersect at two points.

Their coverage regions overlap.

---

Why Overlapping Coverage Matters

Overlap can be important when designing:

• cellular networks
• Wi-Fi systems
• radar networks
• emergency communication systems
• sensor networks

Too little overlap may create gaps.

Too much overlap may be inefficient or cause interference in some systems.

Circle geometry provides a simplified way to investigate these designs.

---

Circular Motion

An object undergoing circular motion remains a constant distance from a central point.

If an object moves around:

center = (2, 3)

at radius:

r = 5 m

then every position on its circular path satisfies:

(x − 2)² + (y − 3)² = 25

The object's coordinates change as it moves, but its distance from the center remains constant.

---

Position in Circular Motion

Suppose a rotating object follows:

x² + y² = 100

The center is:

(0, 0)

and the radius is:

10 m

Possible positions include:

(10, 0)

(0, 10)

(−10, 0)

(0, −10)

All satisfy:

x² + y² = 100

Other points on the circle also represent possible positions.

---

Verifying a Position

Suppose an object follows:

x² + y² = 100

Could it be at:

(6, 8)?

Substitute:

6² + 8²

= 36 + 64

= 100

Yes.

Therefore:

(6, 8) lies on the circular path

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Radius and Direction in Circular Motion

At any instant, a radius can be drawn from the center to the moving object.

The radius indicates the object's radial direction.

For uniform circular motion, the object's instantaneous velocity is directed:

tangent to the circle

Therefore, the velocity direction is:

perpendicular to the radius

This connects circle geometry directly with physics.

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Tangent Direction in Circular Motion

Suppose a particle is at point P on a circular path with center O.

The radius is:

OP

The tangent at P is perpendicular to OP.

The instantaneous velocity of the particle points along this tangent.

Therefore:

radius ⟂ instantaneous velocity

for ideal circular motion.

This geometric relationship is fundamental in mechanics.

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Centripetal Acceleration

Although the velocity is tangent to the path, the acceleration of an object in uniform circular motion points:

toward the center

This acceleration is called:

centripetal acceleration

Therefore:

• radius points between center and object
• velocity is tangent to the circle
• centripetal acceleration points inward toward the center

Circle geometry helps describe these directions clearly.

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Wheels and Rotating Machinery

Wheels, gears, turbines, flywheels, and pulleys involve circular geometry.

Important quantities include:

• radius
• diameter
• center
• circumference
• tangent direction

A point on the edge of a rotating wheel follows a circular path relative to the wheel's center.

The wheel's radius therefore determines the size of that path.

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Tangents and Belt Systems

A belt passing around a circular pulley touches the pulley tangentially.

At the point where a straight section of belt leaves the pulley:

the belt is tangent to the circle

The radius to that contact point is perpendicular to the belt.

Engineers can use this property when designing:

• belt drives
• pulley systems
• conveyor systems
• mechanical transmissions

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Ferris Wheels

A Ferris wheel can be modeled as a circle.

Suppose the wheel has:

radius = 20 m

and its center is located:

22 m above the ground.

If we choose the center's horizontal position as x = 0, the center is:

(0, 22)

The wheel can be modeled by:

x² + (y − 22)² = 400

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Highest and Lowest Points of a Ferris Wheel

For the Ferris wheel:

center height = 22 m

radius = 20 m

Highest point:

22 + 20 = 42 m

Lowest point:

22 − 20 = 2 m

Circle geometry therefore tells us immediately the maximum and minimum heights.

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Checking a Ferris Wheel Position

For:

x² + (y − 22)² = 400

suppose a passenger car is at:

(16, 10)

Check:

16² + (10 − 22)²

= 256 + (−12)²

= 256 + 144

= 400

Therefore:

(16, 10) lies on the wheel

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Circular Tracks

A circular running track can be represented using a circle equation.

Suppose its center is:

(50, 40)

and radius:

30 m

The central path could be modeled as:

(x − 50)² + (y − 40)² = 900

A runner moving around the track remains approximately 30 m from the center.

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Distance Around a Circular Path

A circle equation describes location, but sometimes we need the distance traveled around the circle.

The circumference is:

C = 2πr

If:

r = 30 m

then:

C = 2π(30)

C = 60π m

Approximately:

C ≈ 188.5 m

One complete lap would therefore be about 188.5 m.

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Multiple Laps

If a runner completes:

5 laps

of a circular path with circumference:

60π m

then:

distance = 5(60π)

distance = 300π m

Approximately:

942.5 m

This connects coordinate geometry with measurement and motion.

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Circular Roads and Roundabouts

A roundabout can be modeled using a circle.

Suppose a circular roundabout has:

center = (10, 15)

radius = 8 m

Its outer boundary can be represented by:

(x − 10)² + (y − 15)² = 64

Roads entering and leaving the roundabout may be modeled using tangent lines.

A tangent road meets the circular boundary smoothly at a point of contact.

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Why Tangents Matter in Road Design

At a tangent point:

radius ⟂ tangent

This allows designers to determine the direction of a straight road meeting a circular curve.

In real transportation engineering, road transitions are more complicated than ideal circle-and-line models, but tangent geometry provides an important mathematical foundation.

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Circular Buildings and Architecture

Architects may use circular geometry when designing:

• domes
• arenas
• plazas
• towers
• windows
• arches
• circular rooms

A coordinate model can help determine:

• boundaries
• dimensions
• positions
• distances
• clearances
• intersections with other structures

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Example: Circular Plaza

A circular plaza has:

center = (30, 20)

radius = 15 m

Its boundary is:

(x − 30)² + (y − 20)² = 225

Suppose a lamp is planned at:

(39, 32)

Calculate:

(39 − 30)² + (32 − 20)²

= 9² + 12²

= 81 + 144

= 225

Therefore:

the lamp lies exactly on the plaza boundary

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Straight Paths Through Circular Areas

A straight path through a circular area forms a:

chord

Suppose a circular park has radius:

13 m

A straight path passes:

5 m from the center

The perpendicular from the center bisects the chord.

Let half the path inside the park equal x.

Then:

x² + 5² = 13²

x² + 25 = 169

x² = 144

x = 12

Therefore:

full path length = 2(12)

24 m

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Why the Chord Calculation Works

The perpendicular from the center to a chord:

bisects the chord

This creates two right triangles.

Each triangle has:

• hypotenuse = radius
• one leg = distance from center to path
• other leg = half the chord

This allows the Pythagorean theorem to be used.

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General Chord Application

If a circular region has radius r and a straight path passes a perpendicular distance d from its center, the length L inside the circle is:

L = 2√(r² − d²)

For example:

r = 10

d = 8

Then:

L = 2√(10² − 8²)

L = 2√(100 − 64)

L = 2√36

L = 12

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Manufacturing and Circular Components

Circle geometry is important when manufacturing:

• bearings
• pipes
• washers
• gears
• wheels
• discs
• seals
• circular openings

Manufacturers must control quantities such as:

• center position
• radius
• diameter
• spacing
• clearance

Coordinate geometry can help determine whether parts will fit together correctly.

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Clearance Between Circular Objects

Suppose two circular components have:

Component A:

center = (0, 0)

radius = 4 cm

Component B:

center = (10, 0)

radius = 3 cm

Distance between centers:

10 cm

Combined radii:

4 + 3 = 7 cm

Since:

10 > 7

there is a gap.

Gap between their nearest edges:

10 − 7

gap = 3 cm

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When Circular Objects Touch

Two circular objects touch externally when:

distance between centers = sum of radii

Suppose:

r₁ = 6 cm

r₂ = 4 cm

If the centers are:

10 cm apart

then:

6 + 4 = 10

Therefore, the circles:

touch externally at exactly one point

This principle is useful when analyzing gears, rollers, and other circular components.

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When Circular Objects Overlap

If:

distance between centers < sum of radii

the circular regions may overlap.

However, if one circle is much smaller and lies completely inside the other, we must also compare the distance with:

|r₁ − r₂|

For two circle boundaries to cross at two points:

|r₁ − r₂| < d < r₁ + r₂

where d is the distance between centers.

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Robotics and Path Planning

Robots may use circular models for:

• turning paths
• sensor ranges
• obstacle boundaries
• safety distances

Suppose a robot must remain at least:

5 m

from an obstacle located at:

(8, 6)

The safety boundary can be modeled as:

(x − 8)² + (y − 6)² = 25

The robot should remain outside this circle.

Therefore, safe positions satisfy:

(x − 8)² + (y − 6)² > 25

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Computer Graphics and Games

Circle equations are also useful in computer graphics and game development.

A game may approximate an object using a circular boundary.

Suppose:

Player center = (5, 4)

Player radius = 1

Obstacle center = (9, 7)

Obstacle radius = 2

The program can calculate the distance between centers and compare it with:

1 + 2 = 3

If the center distance is less than or equal to 3, the circular boundaries touch or overlap.

This provides a simple form of collision detection.

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Worked Collision Example

Player:

center = (2, 3)

radius = 2

Obstacle:

center = (5, 7)

radius = 3

Distance between centers:

D = √[(5 − 2)² + (7 − 3)²]

D = √(3² + 4²)

D = 5

Combined radii:

2 + 3 = 5

Therefore:

the two circular boundaries are touching

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Medical and Scientific Imaging

Circular geometry can also appear in:

• microscope images
• medical scans
• particle detection
• astronomy images
• image-processing systems

A circular feature can be described by its:

• center coordinates
• radius
• diameter
• boundary equation

Software can then measure or compare these features.

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Circular Cross-Sections

Many three-dimensional objects have circular cross-sections.

Examples include:

• pipes
• cylinders
• cables
• rods
• blood vessels
• tunnels

If a cross-section is approximately circular, its geometry can be analyzed using circle equations and radius measurements.

The actual three-dimensional object is not a circle, but a two-dimensional slice through it may be.

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Science Example: Circular Particle Path

In some simplified physics situations, a charged particle may follow a circular path.

Suppose its path has:

center = (−2, 3)

radius = 6 cm

The path is modeled by:

(x + 2)² + (y − 3)² = 36

A measured particle position can be tested against this equation to determine whether it is consistent with the predicted circular path.

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Engineering Example: Rotating Arm

Suppose the end of a rotating robotic arm remains:

1.5 m

from its pivot.

If the pivot is represented by:

(4, 2)

then the endpoint follows:

(x − 4)² + (y − 2)² = 2.25

because:

1.5² = 2.25

The equation describes all possible endpoint positions during a complete rotation.

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Design Constraints

Engineers often need to determine whether a circular object fits inside a particular region.

Suppose a circular component has:

center = (8, 6)

radius = 3 cm

A rectangular housing extends from:

x = 0 to x = 12

and:

y = 0 to y = 10

The circle extends horizontally from:

8 − 3 = 5

to:

8 + 3 = 11

and vertically from:

6 − 3 = 3

to:

6 + 3 = 9

All of these values lie inside the housing boundaries.

Therefore, the circular component fits within the rectangular region.

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Checking Clearance from a Wall

Suppose a circular wheel has:

center = (6, 5)

radius = 4

A vertical wall lies along:

x = 11

The nearest point of the wheel to the wall has:

x = 6 + 4

x = 10

Therefore, the clearance is:

11 − 10

1 unit

This type of calculation is useful in design and manufacturing.

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Scale Matters

Suppose a map uses:

1 coordinate unit = 5 km

A circular region has radius:

8 coordinate units

The actual radius is:

8 × 5

40 km

The actual diameter is:

80 km

Always interpret coordinate measurements using the scale of the model.

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Units Matter

Circle equations may use coordinates measured in:

• metres
• kilometres
• centimetres
• pixels
• map units

The radius has the same units as the coordinate axes.

However, r² in the equation has squared units mathematically.

For example, if:

r = 5 m

then:

r² = 25 m²

Even though the equation contains r², the physical radius remains:

5 m

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Accuracy and Idealization

Real circular objects may not be perfect circles.

For example:

• a tire deforms under load
• an orbit may be elliptical
• radar range may vary by direction
• a roundabout may include noncircular transition curves
• manufactured components have tolerances

Circle geometry provides a mathematical approximation.

A good model should be judged by whether it is sufficiently accurate for its intended purpose.

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A Complete Modeling Example

A rotating observation ride has:

center = (12, 25)

radius = 20 m

Its path is:

(x − 12)² + (y − 25)² = 400

The highest point is:

25 + 20

45 m

The lowest point is:

25 − 20

5 m

The leftmost point is:

12 − 20

x = −8

The rightmost point is:

12 + 20

x = 32

The equation therefore provides both the ride's position and dimensions.

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A Complete Distance Example

A radar station is located at:

(10, 20)

with maximum idealized range:

25 km

An aircraft is located at:

(25, 40)

Calculate squared distance:

D² = (25 − 10)² + (40 − 20)²

D² = 15² + 20²

D² = 225 + 400

D² = 625

Radar range squared:

r² = 25²

r² = 625

Therefore:

the aircraft lies exactly on the modeled maximum-range boundary

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A Complete Design Example

Two circular gears have:

Gear A:

center = (0, 0)

radius = 6 cm

Gear B:

center = (10, 0)

radius = 4 cm

Distance between centers:

d = 10 cm

Sum of radii:

6 + 4 = 10 cm

Therefore:

the two circular gear boundaries touch externally

This is the geometric condition needed for two idealized circles to meet at exactly one point.

Real gear teeth make actual gear design considerably more complex, but the circle model provides a useful starting point.

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A Reliable Strategy for Real-World Circle Problems

When solving an applied circle problem:

1. Identify what the coordinates represent.
2. Identify the scale and units.
3. Determine the center.
4. Determine the radius or diameter.
5. Write the circle equation if needed.
6. Decide whether you are modeling the boundary or the entire region.
7. Identify the quantity being asked for.
8. Use distance, radius, chord, tangent, or circumference relationships as appropriate.
9. Calculate carefully.
10. Include appropriate units.
11. Interpret the mathematical result in context.
12. Consider whether the circle is an exact representation or an idealized model.

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Choosing the Right Circle Tool

Different problems require different ideas.

Recognizing the type of problem is often more important than memorizing additional formulas.

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Common Mistakes

Confusing the boundary with the entire region

An equation with = describes the circle itself. An inequality with ≤ can describe the interior and boundary.

Using r instead of r² in the equation

A radius of 8 gives:

r² = 64

Reading center signs incorrectly

In:

(x + 4)² + (y − 3)² = 25

the center is:

(−4, 3)

Ignoring coordinate scale

A radius of 5 coordinate units does not necessarily mean 5 metres.

Forgetting units

A numerical result is incomplete in many applied problems without its physical unit.

Calculating a square root unnecessarily

For inside/on/outside tests, comparing D² with r² is often sufficient.

Assuming circular models are always exact

Real systems may differ from ideal mathematical circles.

Using circumference when straight-line distance is required

Distance around a circle and distance directly between two points are different quantities.

Using diameter instead of radius

Circle equations use the radius.

Assuming all circular motion has constant speed

A circular path describes the shape of the motion. It does not by itself tell us whether speed is constant.

Assuming circular coverage is realistic in every direction

Real signals and sensors may be affected by environmental conditions.

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Key Terms

Circle model — mathematical representation of a circular object, path, or boundary.

Center — fixed point determining the location of a circle.

Radius — constant distance from the center to the circular boundary.

Diameter — distance across a circle through its center; equal to 2r.

Circle equation — coordinate relationship (x − h)² + (y − k)² = r².

Boundary — points exactly r units from the center.

Circular region — area consisting of points inside a circular boundary.

Squared distance — distance comparison made before taking a square root.

Circular motion — motion along a circular path at a constant distance from a center.

Tangent — line touching a circle at exactly one point.

Chord — segment joining two points on a circle.

Circumference — distance around a circle.

Coverage area — region modeled as being within the operating range of a transmitter, sensor, or similar device.

Collision detection — process of determining whether modeled objects touch or overlap.

Clearance — distance separating an object from another object or boundary.

Scale — relationship between coordinate units and actual physical distances.

Mathematical model — mathematical representation used to describe or predict a real system.

Idealization — simplified representation that deliberately ignores some real-world complexity.

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Key Takeaways

• Circle geometry can model circular objects, boundaries, paths, and motion.
• The equation (x − h)² + (y − k)² = r² describes a circular boundary.
• The center determines the circle's location, while the radius determines its size.
• Inequalities can describe the region inside or outside a circular boundary.
• Comparing squared distance with r² provides an efficient way to determine whether a location is inside, on, or outside a circle.
• Coordinate scales must be considered when converting mathematical distances to real measurements.
• Circular models are useful for radar, communications, sensors, robotics, transportation, architecture, and manufacturing.
• An object in circular motion remains a constant distance from the center.
• Instantaneous velocity in circular motion is tangent to the path and perpendicular to the radius.
• A straight path through a circular region forms a chord.
• Tangent relationships are useful when straight paths meet circular objects.
• Circumference determines the distance traveled during a complete revolution around a circular path.
• Distances between centers and radii can determine whether circular objects are separate, touching, or overlapping.
• Circle geometry can be used for collision detection and clearance calculations.
• Real-world circular models often involve assumptions and simplifications.
• A useful mathematical solution should include both a calculation and an interpretation of what the result means in the real situation.

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Check Your Understanding

1. A circular fountain has center (6, 8) and radius 5 m. Write an equation representing its boundary.
2. Explain the difference between (x − 4)² + (y − 3)² = 25 and (x − 4)² + (y − 3)² ≤ 25.
3. A radar station at (10, 15) has an idealized range of 20 km. Write an equation for its maximum-range boundary.
4. For the radar system in Question 3, determine whether an object at (22, 31) is inside, on, or outside the detection range.
5. A circular track has radius 35 m. Calculate the distance traveled during one complete lap.
6. A runner completes 8 laps of the track in Question 5. Calculate the total distance traveled.
7. A Ferris wheel has radius 18 m and its center is 20 m above the ground. Determine its highest and lowest heights.
8. A circular park has radius 10 m. A straight path passes 6 m from its center. Calculate the length of the path inside the park.
9. Two circular communication zones have radii 8 km and 5 km. Their centers are 10 km apart. Determine whether their boundaries are separate, tangent, or intersecting.
10. Two circular machine components have radii 4 cm and 3 cm, and their centers are 11 cm apart. Calculate the clearance between their nearest edges.
11. A rotating robotic arm has a length of 2.5 m and pivots around (4, 6). Write an equation describing the path of its endpoint.
12. A particle is expected to follow (x − 2)² + (y + 3)² = 100. Determine whether the measured position (8, 5) lies on the predicted path.
13. Explain why comparing squared distances rather than calculating actual distances can be useful in computer collision-detection systems.
14. A designer uses a perfect circular model for the coverage of a wireless transmitter. Give two reasons why the real coverage area might differ from the mathematical model.
15. A circular safety zone has center (12, 9) and radius 7 m. A robot travels through positions (15, 13), (18, 13), and (20, 15). Determine whether each position is inside, on, or outside the safety zone, and explain how this information could be used in automated path planning.