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## Key Concepts

• Draw tangents to a circle
• Identify special segments and lines.
• Find lengths in a coordinate plane.
• Draw common tangents.
• Verify a tangent to a circle.
• Find the radius of a circle.
• Tangents from a common external point are congruent.

## Tangents

Question:

How are the lengths of tangent segments related?

A line can intersect a circle at 0, 1, or 2 points. If a line is in the plane of a circle and intersects the circle at 1 point, the line is a tangent.

Step 1:

Draw tangents to a circle

Draw a circle. Use a compass to draw a circle. Label the center P.

Draw tangents

Draw lines and BC so that they intersect  ⊙  P only at A and C, respectively. These lines are called tangents.

Measure segments

AB↔  and BC↔ are called tangent segments. Measure and compare the lengths of the tangent segments.

A circle is the set of all points in a plane that are equidistant from a given point called the center of the circle.

A circle with center P is called “circle P” and can be written ⊙  P.

A segment whose endpoints are the center and any point on the circle is a radius

A chord is a segment whose endpoints are on a circle.

A diameter is a chord that contains the center of the circle.

A secant is a line that intersects a circle in two points.

A tangent is a line in the plane of a circle that intersects the circle in exactly one point, the point of tangency.

The tangent ray

AB and the tangent segment AB are also called tangents

Example 1:

Tell whether the line, ray, or segment is best described as a radius, chord, diameter, secant, or tangent of ⊙C.

a. AC

b. AB

c.  DE

d.  AE

Solution

1. AC is a radius because C is the center and A is a point on the circle.
2. AB is a diameter because it is a chord that contains the center C.
3. DE  is a tangent ray because it is contained in a line that intersects the circle at only one point.
4. AE is a secant because it is a line that intersects the circle in two points.

### Guided Practice

Example 1

1. In Example 1, what word best describes AG and CB ?
2. In Example 1, name a tangent and a tangent segment.

Solution:

1. AG = Chord/Diameter, CB =radius,
2. DE = Tangent and DE

The words radius and diameter are used for lengths as well as segments.

For a given circle, think of a radius and a diameter as segments and the radius and the diameter as lengths.

### Find lengths in circles in a coordinate plane

Use the diagram to find the given lengths.

2. Diameter of A
4. Diameter of B

Solution: Count squares

1. The radius of ⨀A is 3 units.
2. The diameter of ⨀A is 6 units.
3. The radius of ⨀B is 2 units.
4. The diameter of ⨀B is 4 units.

### Guided Practice

Example 2:

Use the diagram in Example 2 to find the radius and diameter of ⨀C and ⨀D.

Solution:

The radius of ⨀C is 3 units.

The diameter of ⨀C is 6 units

The radius of ⨀D is 2 units.

The radius of ⨀D is 4 units.

### Coplanar Circles

Two circles can intersect in two points, one point, or no points.

Coplanar circles that intersect in one point are called tangent circles.

Coplanar circles that have a common center are called concentric circles.

### Common Tangents

A line, ray, or segment that is tangent to two coplanar circles is called a common tangent.

Draw common tangents

### Guided Practice

Example 3

Tell how many common tangents the circles have and draw them.

### Theorem 10.1

In a plane, a line is tangent to a circle if and only if the line is perpendicular to a radius of the circle at its endpoint on the circle.

GIVEN: Line m is tangent to ⨀Q at P.

PROVE c: m ⊥ QP

1. Assume m is not perpendicular to QP

Then the perpendicular segment from Q to m intersects m at some other point R.

Because m is a tangent, R cannot be inside ⨀Q. Compare the length QR to QP.

1. Because QR is the perpendicular segment from Q to m. QR is the shortest segment from Q to m. Now compare QR to QP.
1. Use your results from parts (a) and (b) to complete the indirect proof.

Verify a tangent to a circle

EXAMPLE 4

In the diagram

PT is a radius of ⊙P. Is ST tangent to ⨀P?

Solution:

Use the Converse of the Pythagorean Theorem.

Because 122 + 352= 372 △PST is a right triangle and ST ⊥ PT

So, ST is perpendicular to a radius of ⨀P at its endpoint on ⨀P.

By Theorem 10.1,

ST is tangent to ⨀P.

EXAMPLE 5:

In the diagram, B is a point of tangency. Find the radius r of ⨀C.

You know from Theorem 10.1 that

AB ⊥ BC , so △ABC is a right triangle.

You can use the Pythagorean Theorem.

AC2 =BC 2 +AB 2                         (Pythagorean Theorem)

(r + 50)2 = r 2 + 802                            (Substitute)

r 2 + 100r + 2500 = r 2 + 6400               (Multiply)

100r = 3900                                    (Subtract from each side)

∴   r =39 ft                                   (Divide each side by 100)

### Theorem 10.2

Tangent segments from a common external point are congruent.

Proving Theorem10.2

Write proof that tangent segments from a common external point are congruent.

GIVEN:

SR and ST are tangent to ⊙P.

PROVE: SR > ST

Plan for Proof

Use the hypotenuse–leg congruence

Theorem to show that

△ SRP △STP

Join P, A; PB and PS

∠PRS=∠STP= 900

Now, the right triangles

∆PRS and ∆PTS

PR=PT (radii of the same circle)

PS=PS (Common)

Therefore, by R.H.S congruency axiom

∆PRS ≅ ∆PTS

This gives SR ≅ ST

Hence proved.

Example 6

RS is tangent to ⨀C at S and RT is tangent to ⨀C at T. Find the value of x.

Solution:

RT = RS                  (Tangent segments from the same point are ≅)

3x+4 =28               (Substitute)

3x =28-4

3x=24

x=8                        (solve for x)

### Problem solving

#### GLOBAL POSITIONING SYSTEM (GPS)

GPS satellites orbit about 11,000 miles above Earth. The mean radius of Earth is about 3959 miles. Because GPS signals cannot travel through Earth, a satellite can transmit signals only as far as points A and C from point B, as shown. Find BA and BC to the nearest mile.

Solution:

ED =3959 mi     is the radius of Earth

DB=1100 mi

AB=?     BC=?          AB, BC are tangents

AB=BC (∵ Tangent segments from a common external point are congruent.)

BE=ED+DB=3959+11000=14959 mi

AE ⊥ AB The tangent at any point of a circle is perpendicular to the radius through the point of contact.

△EAB is right angled triangle. By Pythagoras theorem

AE2 +AB2= BE2

AB2 = BE2– AE2

= 149592 – 39592

= (14959+3959) (14959–3959)

=11000 x 18918

AB= √18918 × 11000

=14425.60 mi

1. Copy the diagram. Tell how many common tangents the circles have and draw.
1. Determine whether AB is to ⊙C. Explain.
1. Find the values of the variable.
1. Find the values of x and y.

1.

2. From △CAB

BC2 = AC2+ AB2 (∵ Pythagoras theorem)

52 = 32+ 42

25=25

So△CAB is right angled triangle.

∠CAB = 900

CA ⊥ AB

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ AB is a tangent

3. Tangent segments from a common external point are congruent.

7x – 6 =3x+10

7x-3x=10+6

4x=16

∴   x=4

Step 1:

Linear pair = 1800

(2x+3)0 + 1370= 1800

x0 + 1400= 1800

2x0 = 400

x0 = 200

(2x+3)0 =2 x 200+3= 430

Step 2:  Two lines are parallel to each other, then

Alternate exterior angles are equal

(4y−7)o = (2x+3)0

(4y−7)o = 430

4y0 = 500

y0 = (252)0

## Exercise

1. Find the value of x and y in the following figure
1. When will two lines tangent to the same circle not intersect?
2. Find the value of the variables in the following figures:
1. Determine whether AB tangent is tangent to circle C. Explain.

### What have we learned

• How to draw tangents to a circle.
• How to Identify special segments and lines.
• How to find lengths in a coordinate plane.
• How to draw common tangents.
• How to verify a tangent to a circle.
• How to find the radius of a circle.
• How tangents from a common external point are congruent

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