4. Tangents and Chords

Tangents and Chords

Circles contain several important types of lines and line segments. Two of the most useful are tangents and chords.

A chord joins two points on a circle.

A tangent touches a circle at exactly one point.

These simple definitions lead to important geometric relationships involving radii, right angles, distances, and coordinate geometry.

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Chords

A chord is a line segment with both endpoints on a circle.

If points A and B lie on a circle, then segment AB is a chord.

A circle can contain infinitely many different chords.

Chords can:

• have different lengths
• appear at different positions
• pass near the center
• pass through the center
• be horizontal, vertical, or diagonal

Their defining property is simply that:

both endpoints lie on the circle

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The Diameter Is a Special Chord

A diameter is a chord that passes through the center of the circle.

Therefore:

every diameter is a chord

but:

not every chord is a diameter

The diameter is also the:

longest possible chord of a circle

If the radius is r, then:

diameter = 2r

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Why the Diameter Is the Longest Chord

Consider chords at different distances from the center.

A chord passing through the center stretches across the widest part of the circle.

As a chord moves farther away from the center, it becomes shorter.

Therefore:

the closer a chord is to the center, the longer it is

The chord passing directly through the center is the diameter and has maximum length:

2r

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Equal Chords

In the same circle, or in congruent circles:

equal chords are the same distance from the center

The reverse is also true:

chords that are the same distance from the center have equal lengths

This creates an important connection between:

• chord length
• distance from the center

For example, if chords AB and CD are each 4 cm from the center, then:

AB = CD

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Perpendicular from the Center to a Chord

Suppose a perpendicular segment is drawn from the center of a circle to a chord.

An important circle theorem states:

A perpendicular from the center of a circle to a chord bisects the chord.

To bisect means to divide into two equal parts.

If OM is perpendicular to chord AB, then:

AM = MB

and M is the midpoint of AB.

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Why a Perpendicular Radius Bisects a Chord

Suppose O is the center, AB is a chord, and OM is perpendicular to AB.

Because OA and OB are radii:

OA = OB

Triangles OMA and OMB are right triangles.

They share side OM.

Their hypotenuses OA and OB are equal.

Therefore, the two triangles are congruent, so:

AM = MB

Thus the perpendicular from the center bisects the chord.

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Finding a Chord Length

Suppose a circle has:

radius = 10 cm

and a chord is:

6 cm from the center.

The perpendicular from the center bisects the chord.

Let half the chord be x.

A right triangle is formed with:

• hypotenuse = 10
• one leg = 6
• other leg = x

Using the Pythagorean theorem:

x² + 6² = 10²

x² + 36 = 100

x² = 64

x = 8

This is only half the chord.

Therefore:

chord length = 2(8)

chord length = 16 cm

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General Chord-Length Relationship

If:

• r = radius
• d = perpendicular distance from the center to the chord
• L = chord length

then half the chord has length:

L/2

Using the Pythagorean theorem:

(L/2)² + d² = r²

Therefore:

L/2 = √(r² − d²)

and:

L = 2√(r² − d²)

This equation shows directly why chords become shorter as their distance from the center increases.

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Tangents

A tangent is a line that touches a circle at exactly one point.

The point where the tangent touches the circle is called the:

point of tangency

If line l touches a circle only at T, then:

• l is tangent to the circle
• T is the point of tangency

Unlike a secant, a tangent does not pass through the interior of the circle.

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Radius and Tangent Relationship

One of the most important properties of a circle is:

A tangent to a circle is perpendicular to the radius drawn to the point of tangency.

If:

O = center

T = point of tangency

then:

OT ⟂ tangent

Therefore, the angle between the radius and tangent is:

90°

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Why the Tangent Is Perpendicular to the Radius

Imagine a circle with center O and a tangent touching it at T.

The distance OT is exactly:

r

Now consider any other point P on the tangent.

If OP were also equal to or less than r, then P would lie on or inside the circle.

But a tangent touches the circle only at T.

Therefore, every other point on the tangent must be farther from O than T is.

This means OT is the:

shortest distance from the center to the tangent line

The shortest distance from a point to a line is always measured along a perpendicular.

Therefore:

OT is perpendicular to the tangent

and:

angle OTP = 90°

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The Converse Tangent Theorem

The relationship also works in reverse.

If:

• T lies on a circle
• OT is a radius
• a line through T is perpendicular to OT

then that line is:

tangent to the circle at T

This is useful when proving that a particular line is tangent.

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Tangent vs Secant

A tangent intersects a circle at:

one point

A secant intersects a circle at:

two points

A chord is the segment joining the two points where a secant crosses the circle.

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Two Tangents from the Same External Point

Suppose point P lies outside a circle.

Two tangents are drawn from P, touching the circle at A and B.

An important theorem states:

Tangent segments drawn from the same external point are equal.

Therefore:

PA = PB

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Why the Two Tangent Segments Are Equal

Suppose O is the center.

Because OA and OB are radii:

OA = OB

Because radii are perpendicular to tangents:

OA ⟂ PA

and:

OB ⟂ PB

Therefore, triangles OAP and OBP are right triangles.

They share:

OP

and have equal legs:

OA = OB

The triangles are congruent.

Therefore:

PA = PB

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Worked Tangent-Length Example

A point P lies outside a circle.

Two tangent segments are drawn:

PA = 3x + 2

PB = 5x − 8

Because tangent segments from the same external point are equal:

3x + 2 = 5x − 8

Add 8:

3x + 10 = 5x

10 = 2x

x = 5

Then:

PA = 3(5) + 2

PA = 17

and:

PB = 5(5) − 8

PB = 17

Therefore:

PA = PB = 17

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Finding Tangent Length Using the Pythagorean Theorem

Suppose:

• O is the center
• T is the point of tangency
• P is an external point

Because OT is perpendicular to PT, triangle OTP is a right triangle.

Suppose:

OP = 13 cm

OT = 5 cm

Find PT.

Use:

OP² = OT² + PT²

13² = 5² + PT²

169 = 25 + PT²

PT² = 144

PT = 12 cm

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Tangents in Coordinate Geometry

The tangent-radius relationship becomes especially useful on a coordinate plane.

If we know:

• the center of a circle
• the point of tangency

we can calculate the slope of the radius.

Then we find the slope of the tangent using the fact that perpendicular slopes are:

negative reciprocals

This allows us to write the equation of the tangent line.

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Finding a Tangent Equation

Suppose a circle has center:

O(0, 0)

and the tangent touches the circle at:

T(3, 4)

First find the slope of radius OT:

m_radius = (4 − 0)/(3 − 0)

m_radius = 4/3

The tangent is perpendicular to the radius.

Therefore:

m_tangent = −3/4

Now use point-slope form through T(3, 4):

y − 4 = −3/4(x − 3)

This is the equation of the tangent.

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Converting the Tangent Equation

Start with:

y − 4 = −3/4(x − 3)

Distribute:

y − 4 = −3/4x + 9/4

Add 4:

y = −3/4x + 9/4 + 16/4

Therefore:

y = −3/4x + 25/4

The tangent touches the circle at:

(3, 4)

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Checking the Tangent Relationship

Radius slope:

4/3

Tangent slope:

−3/4

Multiply:

(4/3)(−3/4) = −1

Therefore, the lines are perpendicular.

This confirms the tangent-radius relationship.

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Horizontal Radius and Vertical Tangent

Suppose the center is:

(2, 3)

and the point of tangency is:

(7, 3)

The radius from (2, 3) to (7, 3) is horizontal.

Its slope is:

0

Therefore, the tangent is vertical.

The tangent passes through x = 7.

Its equation is:

x = 7

This special case cannot be handled by simply multiplying slopes because a vertical line has undefined slope.

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Vertical Radius and Horizontal Tangent

Suppose the center is:

(4, −2)

and the point of tangency is:

(4, 5)

The radius is vertical.

Therefore, the tangent is horizontal.

Since it passes through:

(4, 5)

its equation is:

y = 5

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Verifying a Tangent Point on a Circle

Before writing a tangent equation, the proposed point of tangency should actually lie on the circle.

Suppose the circle is:

x² + y² = 25

and the proposed tangent point is:

(3, 4)

Check:

3² + 4² = 25

9 + 16 = 25

25 = 25

Therefore:

(3, 4) lies on the circle

It can be used as a point of tangency.

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Coordinate Example with a Shifted Center

Consider:

(x − 2)² + (y − 1)² = 25

Suppose the tangent point is:

(5, 5)

First check that the point lies on the circle:

(5 − 2)² + (5 − 1)²

= 3² + 4²

= 9 + 16

= 25

Yes.

The center is:

(2, 1)

Slope of radius:

m_radius = (5 − 1)/(5 − 2)

m_radius = 4/3

Therefore:

m_tangent = −3/4

Use point-slope form:

y − 5 = −3/4(x − 5)

This is the tangent equation.

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Finding a Chord Length from Coordinates

Suppose the endpoints of a chord are:

A(−2, 3)

and:

B(4, 3)

Because the y-coordinates are equal, the chord is horizontal.

Its length is simply:

4 − (−2)

AB = 6 units

For diagonal chords, use the distance formula:

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

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Worked Diagonal Chord Example

Suppose:

A(1, 2)

B(7, 10)

Then:

AB = √[(7 − 1)² + (10 − 2)²]

AB = √(6² + 8²)

AB = √(36 + 64)

AB = √100

AB = 10

If both A and B lie on the same circle, AB is a chord of length:

10 units

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Finding the Midpoint of a Chord

Suppose chord AB has endpoints:

A(2, 3)

B(8, 7)

Its midpoint is:

x = (2 + 8)/2 = 5

y = (3 + 7)/2 = 5

Therefore:

midpoint = (5, 5)

If a line from the center passes through this midpoint and is perpendicular to AB, it is the perpendicular bisector of the chord.

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Perpendicular Bisectors and Circle Centers

An important geometric fact is:

The perpendicular bisector of a chord passes through the center of the circle.

This means chords can be used to locate an unknown center.

If we draw:

• the perpendicular bisector of one chord
• the perpendicular bisector of another chord

their intersection gives the:

center of the circle

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Why the Perpendicular Bisector Passes Through the Center

Suppose A and B are endpoints of a chord.

The center O is the same distance from A and B because:

OA = OB

Both are radii.

Any point that is equally distant from A and B lies on the perpendicular bisector of AB.

Therefore:

O lies on the perpendicular bisector of chord AB

This provides a useful geometric method for finding a circle's center.

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Coordinate Geometry: Perpendicular Bisector of a Chord

Suppose a chord has endpoints:

A(2, 2)

B(6, 4)

First find the midpoint:

M = ((2 + 6)/2, (2 + 4)/2)

M = (4, 3)

Find the slope of AB:

m_AB = (4 − 2)/(6 − 2)

m_AB = 2/4

m_AB = 1/2

The perpendicular slope is:

−2

Therefore, the perpendicular bisector has equation:

y − 3 = −2(x − 4)

The center of the circle must lie somewhere on this line.

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Using Two Chords to Find a Center

If the center of a circle is unknown:

1. choose two chords
2. find the midpoint of each chord
3. find each chord's slope
4. determine the perpendicular slope
5. write each perpendicular bisector
6. find where the two bisectors intersect

That intersection is:

the center of the circle

Once the center is known, the radius can be found using the distance from the center to any point on the circle.

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Chord Symmetry

The perpendicular line from the center to a chord creates symmetry.

If OM is perpendicular to chord AB, then:

AM = MB

The two endpoints of the chord are positioned symmetrically around M.

This is why one half of a chord can often be calculated first and then doubled.

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Equal Chords and Central Distance

Suppose two chords in the same circle are both:

12 cm long

Then they are the same perpendicular distance from the center.

Conversely, if two chords are both:

5 cm from the center

then they have the same length.

This property can be used in geometric proofs and calculations.

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Comparing Two Chords

Suppose a circle contains:

Chord A, 3 cm from the center

Chord B, 7 cm from the center

Which chord is longer?

Chord A is closer to the center.

Therefore:

Chord A is longer

No calculation is required.

If two chords are equally distant from the center, they are equal in length.

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Maximum Chord Length

Suppose a circle has:

radius = 9 cm

The longest possible chord is the diameter.

Therefore:

maximum chord length = 2r

maximum chord length = 2(9)

maximum chord length = 18 cm

Any other chord must be shorter than 18 cm.

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Geometric Reasoning with Tangents

Suppose O is the center and T is a point on the circle.

A line passes through T.

If you can prove that:

OT ⟂ line

then you can conclude:

the line is tangent to the circle at T

This converse relationship is often useful in proofs.

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Geometric Reasoning with Chords

Suppose O is the center and AB is a chord.

A line from O meets AB at M.

If:

OM ⟂ AB

then:

AM = MB

If instead you know:

AM = MB

then the line from the center to M is perpendicular to the chord.

These relationships connect:

• midpoint
• perpendicularity
• chord geometry
• circle centers

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A Complete Tangent Problem

A circle has equation:

(x − 1)² + (y − 2)² = 25

Point T is:

(4, 6)

Find the equation of the tangent at T.

First verify T:

(4 − 1)² + (6 − 2)²

= 3² + 4²

= 25

Therefore, T lies on the circle.

Center:

O(1, 2)

Radius slope:

m_radius = (6 − 2)/(4 − 1)

m_radius = 4/3

Tangent slope:

m_tangent = −3/4

Use point-slope form:

y − 6 = −3/4(x − 4)

This is the tangent equation.

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A Complete Chord Problem

A circle has radius:

13 cm

A chord lies:

5 cm

from the center.

A perpendicular from the center bisects the chord.

Let half the chord equal x.

Then:

x² + 5² = 13²

x² + 25 = 169

x² = 144

x = 12

The entire chord is:

2(12)

24 cm

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Real-World Tangents

Tangents appear frequently in real-world designs.

Examples include:

• a wheel touching a road
• a belt touching a pulley
• a straight road meeting a circular roundabout
• mechanical rollers touching surfaces
• gears and drive systems
• circular tracks meeting straight sections

The radius to the point of contact is perpendicular to the tangent direction.

This relationship is useful in engineering and design.

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Real-World Chords

Chords also occur in practical situations.

Examples include:

• bridge beams spanning circular structures
• straight paths crossing circular parks
• supports inside circular frames
• straight cuts across circular materials
• window designs
• circular architectural structures

A chord does not need to pass through the center.

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Designing a Straight Path Across a Circular Park

Suppose a circular park has radius:

25 m

A straight path passes through the park:

7 m from the center

The path inside the park forms a chord.

Half the chord has length x.

Use:

x² + 7² = 25²

x² + 49 = 625

x² = 576

x = 24

Therefore, the complete path across the circular region is:

2(24)

48 m

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Tangent Length and Distance from the Center

Suppose an external point P is:

17 cm

from the center of a circle with radius:

8 cm

Let tangent length PT = x.

Because the radius and tangent are perpendicular:

8² + x² = 17²

64 + x² = 289

x² = 225

x = 15 cm

The tangent segment is 15 cm long.

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A General Tangent-Length Relationship

If:

• d = distance from an external point to the center
• r = radius
• t = tangent length

then:

d² = r² + t²

Therefore:

t = √(d² − r²)

This only applies when the point is outside the circle:

d > r

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Connecting Tangents, Chords, and Right Triangles

Many tangent and chord problems reduce to right triangles.

For tangent problems:

• radius is perpendicular to tangent
• this creates a right triangle

For chord problems:

• perpendicular from center bisects chord
• this creates two right triangles

The Pythagorean theorem is therefore one of the most useful tools in circle geometry.

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A Reliable Strategy for Tangent Problems

When solving a tangent problem:

1. Identify the center.
2. Identify the point of tangency.
3. Draw or identify the radius to that point.
4. Use the fact that radius and tangent are perpendicular.
5. Look for a right triangle if lengths are involved.
6. Use perpendicular slopes if coordinates are involved.
7. Use point-slope form to write a tangent equation if needed.
8. Check that the point of tangency actually lies on the circle.
9. State the result with appropriate units.

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A Reliable Strategy for Chord Problems

When solving a chord problem:

1. Identify the center and chord.
2. Look for a perpendicular from the center to the chord.
3. Recognize that it bisects the chord.
4. Work with half the chord.
5. Identify the radius as the hypotenuse.
6. Apply the Pythagorean theorem.
7. Double the half-chord if the full chord is required.
8. Check that the chord is no longer than the diameter.

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A Reliable Strategy for Coordinate Problems

For a tangent:

1. identify the center
2. verify the point of tangency
3. calculate the radius slope
4. find its negative reciprocal
5. use the tangent point in point-slope form

For a chord:

1. identify the chord endpoints
2. calculate its length if required
3. calculate its midpoint
4. calculate its slope
5. use the perpendicular slope when locating the center or perpendicular bisector

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Common Mistakes

Thinking a tangent crosses the circle

A tangent meets the circle at exactly one point.

Thinking every segment joining points on a circle is a diameter

It is a chord. It is only a diameter if it passes through the center.

Forgetting that a diameter is a chord

Every diameter satisfies the definition of a chord.

Assuming every radius is perpendicular to every chord

A radius is perpendicular to a chord when it meets the chord in the appropriate perpendicular relationship. It is not automatically perpendicular to every chord.

Forgetting that a perpendicular from the center bisects a chord

This relationship often turns one difficult problem into two simple right triangles.

Using the whole chord in the right triangle

When the perpendicular from the center bisects the chord, use:

half the chord

as the triangle leg.

Forgetting to double the result

If you calculate half the chord, multiply by 2 to obtain the complete chord.

Using the radius slope as the tangent slope

The tangent is perpendicular, so use the negative reciprocal.

Changing only the sign of the radius slope

For example, the perpendicular slope to 2/3 is:

−3/2

not −2/3.

Forgetting horizontal and vertical special cases

A horizontal radius has a vertical tangent.

A vertical radius has a horizontal tangent.

Assuming a proposed tangent point lies on the circle

Verify it before using the tangent-radius theorem.

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Key Terms

Chord — line segment whose endpoints lie on a circle.

Diameter — chord passing through the center; the longest chord of a circle.

Tangent — line that meets a circle at exactly one point.

Point of tangency — point where a tangent touches a circle.

Secant — line that intersects a circle at two points.

Radius — segment from the center to a point on the circle.

Perpendicular — lines or segments meeting at a 90° angle.

Bisect — divide into two equal parts.

Midpoint — point dividing a segment into two equal segments.

Perpendicular bisector — line passing through a segment's midpoint at 90°.

Equal chords — chords with equal lengths; in the same circle they are equally distant from the center.

Tangent segments — segments from an external point to points of tangency.

External point — point located outside a circle.

Congruent triangles — triangles having the same size and shape.

Negative reciprocal — slope obtained by reversing a fraction and changing its sign.

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Key Takeaways

• A chord joins two points on a circle.
• A diameter is a special chord that passes through the center.
• Every diameter is a chord, but not every chord is a diameter.
• The diameter is the longest chord in a circle.
• Chords become shorter as their perpendicular distance from the center increases.
• Equal chords in the same circle are equally distant from the center.
• Chords equally distant from the center have equal lengths.
• A perpendicular from the center to a chord bisects the chord.
• The perpendicular bisector of a chord passes through the center.
• Perpendicular bisectors of two chords can be used to locate an unknown circle center.
• A tangent touches a circle at exactly one point.
• The radius drawn to the point of tangency is perpendicular to the tangent.
• If a line through a point on a circle is perpendicular to the radius at that point, the line is tangent.
• Two tangent segments from the same external point have equal lengths.
• Tangent problems frequently create right triangles.
• Chord problems frequently create two congruent right triangles.
• The Pythagorean theorem is useful for calculating tangent and chord lengths.
• In coordinate geometry, tangent slopes are perpendicular to radius slopes.
• The negative reciprocal relationship can be used to write tangent equations.
• Horizontal and vertical tangent cases must be treated separately.
• Tangent and chord properties provide powerful tools for geometric reasoning, proofs, coordinate geometry, and real-world design.

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Check Your Understanding

1. Define a chord and a tangent.
2. Explain why every diameter is a chord but not every chord is a diameter.
3. What angle is formed between a tangent and the radius drawn to the point of tangency?
4. Explain why the radius must be perpendicular to the tangent at the point of contact.
5. A circle has radius 8 cm. What is the maximum possible chord length?
6. A circle has radius 10 cm, and a chord is 6 cm from the center. Calculate the chord's length.
7. A circle has radius 13 cm, and a chord is 12 cm long. Calculate the perpendicular distance from the center to the chord.
8. Two chords of the same circle are both 14 cm long. What can you conclude about their distances from the center?
9. Two tangents are drawn from external point P. One tangent has length 3x + 7 and the other has length 5x − 9. Find x and the length of each tangent.
10. A circle has radius 9 cm. An external point is 15 cm from the center. Calculate the length of a tangent segment from the external point.
11. A circle has center (0, 0) and a tangent point (5, 12). Find the slope of the radius and the slope of the tangent.
12. A circle has equation (x − 2)² + (y − 1)² = 25. Show that (5, 5) lies on the circle, then write an equation for the tangent at that point.
13. A chord has endpoints A(2, 3) and B(8, 7). Find its length, midpoint, and slope.
14. Explain how the perpendicular bisectors of two different chords can be used to locate the center of an unknown circle.
15. A circular park has radius 25 m. A straight path crosses the park 15 m from the center. Calculate the length of the path inside the circular boundary and explain which chord property makes the calculation possible.