SECTION 10.5Tangents

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

is tangent to at point .

and are also called tangents.

AB C A

AB AB

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A common tangent is a line, ray, or segment that is tangent to two circles in the same plane. In each figure below, in l is a common tangent of circles F and G.

Example 1:  a) Using the figure below, draw the common tangents. If no common tangent exists, state no common tangent.

These circles have no common tangents. Any tangent of theinner circle will intercept the outer circle in two points.

Example 1:  b) Using the figure below, draw the common tangents. If no common tangent exists, state no common tangent.

These circles have 2 common tangents.

a) is a radius of . Determine whether is tangent

to . Justify your answer.

KL K LM

K

Example 2:

Test to see if ΔKLM is a right triangle.

?202 + 212 = 292 Pythagorean Theorem

841 = 841 Simplify.

is a right triangle with right angle . So,

is perpendicular to radius at point . Therefore, by

Theorem 10.10, is tangent to .

KLM KLM LM

KL L

LM K

b) is a radius of . Determine whether is tangent

to . Justify your answer.

XY X YZ

X

Example 2:

Test to see if ΔXYZ is a right triangle.

?142 + 162 = 212 Pythagorean Theorem

452 ≠ 441 Simplify.

is not a right triangle. So, is not perpendicular to

radius at point . Therefore, by Theorem 10.10, is not

tangent to .

XYZ YZ

XY Y YZ

X

a) In the figure, is tangent to at . Find the value of .WE D W x

Example 3:

By Theorem 10.10, .

So, is a right triangle.

EW WD

DEW

EW 2 + DW 2 = DE 2 Pythagorean Theorem

242 + x 2 = (x + 16)2 EW = 24, DW = x, and DE = x + 16

576 + x 2 = x 2 + 32x + 256 Multiply.

320 = 32x Simplify.

10 = x Divide each side by 32.

b) In the figure, is tangent to at . Find the value of .IK J K x

Example 3:

By Theorem 10.10, .

So, is a right triangle.

JK IK

JKI

JK 2 + KI 2 = IJ 2 Pythagorean Theorem

x2 + 162 = (x + 8)2 JK = x, KI = 16, and JI = x + 8

x 2 + 256 = x 2 + 16x + 64 Multiply.

192 = 16x Simplify.

12 = x Divide each side by 16.

a) and are tangent to . Find the value of .AC BC Z x

Example 4:

AC = BC Tangents from the same exteriorpoint are congruent.

3x + 2 = 4x – 3 Substitution

2 = x – 3 Subtract 3x from each side.

5 = x Add 3 to each side.

b) and are tangent to . Find the value of .MN MP Q x

Example 4:

MN = MP Tangents from the same exteriorpoint are congruent.

5x + 4 = 8x – 17 Substitution

4 = 3x – 17 Subtract 3x from each side.

21 = 3x Add 17 to each side.

7 = x Divide each side by 7.

Circumscribed PolygonsA polygon is circumscribed about a circle if every side of the polygon is tangent to the circle.

Example 5: a) The round cookies are marketed in a triangular package to pique the consumer’s interest. If ∆QRS is circumscribed about ⨀T, find the perimeter of ∆QRS.

Since is circumscribed about , and

are tangent to

, as are , , , and Therefore,

, , and

.

.

RL RM

QRS T

T

QN QL SM

SM SN

QN QL

SN RL RM

So, 2 cm and 10 cm.

By Segment Addition, , so

8 2 or 6 cm. So, = 6 m. c

QN QL

QL

QL

RL

RL R

SM S

L

QR

Q M

N

R R

Find the perimeter of ΔQRS.

perimeter =

8

36 cm

1610 02

RMQRQNSN SM

Example 5: b) A bouncy ball is marketed in a triangular package to pique the consumer’s interest. If ∆ABC is circumscribed about ⨀G, find the perimeter of ∆ABC.

Since is circumscribed about , and

are tangent to

, as are , , , and Therefore,

, , and

.

.

BF BD

ABC G

T

CE CD AE

AE AF

CE CD

AF BF BD

So, 6 cm and 12 cm.

By Segment Addition, , so

16 6 or 10 cm. So, = 10 cm .

CE CD

C CBD

CD

AE A

BD

B CBD B

F

BD F

Find the perimeter of ΔQRS.

perimeter =

16

56

6 101 2

m

1

c

2

CBA ABFE FEC