Lesson 9.2: Tangent Lines
Objectives:

Find the relationship between a radius and a tangent to a circle

Find the relationship between two tangents drawn from the same point

Circumscribe a circle

Find equations of concentric circles
Standards:
MATH.CA.8-12.GEOM.1.0, MATH.CA.8-12.GEOM.2.0, MATH.CA.8-12.GEOM.7.0, MATH.CA.8-12.GEOM.16.0,
MATH.CA.8-12.GEOM.21.0
MATH.NCTM.9-12.GEOM.1.1, MATH.NCTM.9-12.GEOM.1.3, MATH.NCTM.9-12.GEOM.4.1
Introduction
In this section we will discuss several theorems about tangent lines to
tangent
circles and the applications of these theorems to geometry problems.
Remember that a tangent to a circle is a line that intersects the circle at
point of
tangency
exactly one point and that this intersection point is called the point of
tangency.
09.2.1 Theorem 9-1
Theorem 9-1:
Tangent to a Circle
A tangent line is always at right angles to the radius of the circle at the point of tangency
Proof: We will prove this theorem by contradiction.
We start by making a drawing. Line segment AB is a radius of the circle.
A is the center of the circle and B is the point of intersection between the
radius and the tangent line.
B
Assume that the tangent line is not perpendicular to the radius.
C
D
There must be another point C on the tangent line such that AC is
perpendicular to the tangent line. Therefore, in the right triangle ACB, line
A
AB is the hypotenuse and AC is a leg of the triangle. However, this is not
possible because AC > AB. (Note that AC = length of the radius + DC).
Since our assumption led us to a falsehood, this means that our assumption was incorrect. Therefore, the
tangent line must be perpendicular to the radius of the circle.
Since the tangent to a circle and the radius of the circle make a right angle with each other, we can often use
the Pythagorean Theorem in order to find the length of missing line segments.
Example 1: In the figure, CB is tangent to the circle. Find the length of segment CD.
Solution:
Since CB is tangent to the circle, then CB  AB .
C
D
A
This means that ABC is a right triangle and we can apply the
8 in
5 in
Pythagorean Theorem to find the length of line segment AC.
B
AC 2  AB 2  BC 2
AC 2  25  64  89
AC  89  9.43 in
CD = AC – AD = 9.43 in – 5 in = 4.43 in
C
Example 2: Mark is standing at the top of Mt. Whitney which is 14,500 feet
tall. The radius of the Earth is approximately 3960 miles. (There are 5280 feet
in one mile.) How far can Mark see to the horizon?
B
Solution:
We start by drawing the figure to the right.
A
The distance to the horizon is given by the line segment CB.
Let us convert the height of the mountain from feet into miles.
14500 feet 
1 mile
 2.75 miles
5280 feet
Since CB is tangent to the Earth, ABC is a right triangle and we can use the Pythagorean Theorem.
AC 2  CB 2  AB 2
(3960  2.75 ) 2  CB 2  3960 2
CB  147 .6 miles
09.2.2 Theorem 9-2
Theorem 9-2: Converse of a Tangent to a Circle
If a line is perpendicular to the radius of a circle at its outer endpoint, then the line is tangent to the
circle.
Proof: We will prove this theorem by contradiction.
Since line BC is perpendicular to the radius at its outer endpoint it must
B
touch the circle at point B. For this line to be tangent to the circle, it
must only touch the circle at this point and no other.
A
C
Assume that line BC also intersects the circle at point C.
Since both AB and AC are radii of the circle, AB  AC , and ABC is isosceles.
This means that ABC  ACB  90 .
B
It is impossible to have two right angles inside a triangle.
Since we arrived at a falsehood, our assumption must be incorrect.
C
A
We conclude that line BC is tangent to the circle at point B.
M 8 cm
Example 3: Determine whether LM is a tangent to the circle.
L
6 cm
Solution:
10 cm
LM is tangent to the circle if LM  NM .
N
To show that LMN is a right triangle we use the Pythagorean Theorem:
LM 2  MN 2 64  36  100  10 2
The lengths of the sides of the triangle satisfy the Pythagorean Theorem, so LM is perpendicular to MN and
is therefore a tangent to the circle.
09.2.3 Theorem 9-3
Theorem 9-3: Tangent Segments from a Common External Point
If two segments from the same exterior point are tangent to the circle, then they are congruent.
Proof: The figure on the right shows a diagram of the situation.
Given:
A
AC is a tangent to the circle
BC is a tangent to the circle
Prove:
C
AC  BC
Statements
Reasons
AB is tangent to the circle
given
AC  OA
Theorem 9.1
BC is a tangent to the circle
given
BC  OB
Theorem 9.1
OA  OB
radii of same the circle
OC  OC
same line
AOC  BOC
HL congruence
AC  BC
CPCTC
B
A
O
C
B
Example 4: Find the perimeter of the triangle.
C
7 cm
Solution:
E
All sides of the triangle are tangent to the circle.
F
Theorem 9-3 tells us that:
CE  FC  7 cm
FA  AD  8 cm
DB  BE  12 cm
A 8 cm
D
B
12 cm
The perimeter of the triangle = AF + FC + CE + EB + BD + AD
= 8 cm + 7 cm + 7 cm + 12 cm + 12 cm + 8 cm = 54 cm
Example 5: An isosceles right triangle is circumscribed about a circle with diameter equal 24 inches. Find
the hypotenuse of the triangle.
A
Solution:
D
E
Let’s start by making a sketch.
O
x
Since EO and DO are radii of the circle and AC and AB are
tangents to the circle,
C
x
EO  AC and DO  AB
Therefore, quadrilateral ADOE is a square.
Therefore, AE = AD = 12 in
We can find the length of side ED by using the Pythagorean Theorem.
ED 2  AE 2  AD 2
ED 2  144  144  288
ED  12 2 in
Triangles ADE and ABC are both isosceles right triangles, therefore ADE ~ ABC and all the
corresponding sides are proportional:
AE AD ED


AC AB CB
We can find the length of EC by using on of the ratios above:
AE ED
12
12 2



AC CB
12  x
2x
Cross-multiply to obtain:
x
24 in
x
B
24 x  144 2  12 2 x
24 x  12 2 x  144 2
12 ( 2  2 ) x  144 2
x
144 2
12 ( 2  2 )

12 2
2 2
Rationalize:
x
12 2

2 2
2 2 2 2

24 2  24
2
x  12 2  12 in
The hypotenuse, BC = 2x = 24 2  24 in  58 in
09.2.4 Corollary 9-3a
Corollary 9-3a:
The line from an external point through center of the circle bisects the angle formed by the tangent lines
starting at that same external point.
Proof:
Given:
AC is a tangent to the circle
BC is a tangent to the circle
A
O is the center of the circle
Prove:
ACO  BCO
O
C
We use the same figure we used in Theorem 9-3.
AOC  BOC by HL congruence.
Therefore, ACO  BCO .
09.2.5 Concentric Circles
Concentric circles are circles of different radii that share the same
center point.
Example 6: Write the equations of the concentric circles shown in
the graph.
Solution:
( x  3) 2  ( y  2) 2  4
B
( x  3) 2  ( y  2) 2  9
( x  3) 2  ( y  2) 2  16
( x  3) 2  ( y  2) 2  25
Example 7: Determine if the circles given by the equations x 2  10x  y 2  12 y  57  0 and
x 2  10x  y 2  12 y  36  0 are concentric.
Solution:
To find the answer to this question we must re-write the equations of the circles in standard form and find
the center point of each circle.
To re-write in standard form, we complete the square on the x and y terms separately.
First circle:
x 2  10x 
 y 2  12 y 
 57 

x 2  10x  25  y 2  12 y  36  57  25  36
( x  5) 2  ( y  6) 2  57  61
( x  5) 2  ( y  6) 2  4
Center of the first circle is at point (5, 6).
Second circle:
x 2  10x 
 y 2  12 y 
 36 

x 2  10x  25  y 2  12 y  36  36  25  36
( x  5) 2  ( y  6) 2  36  61
( x  5) 2  ( y  6) 2  25
Center of the first circle is at point (5, 6).
Example 8: Show that the line y = 5 − 2x is tangent to the circle x2 + y2 = 5. Find an equation for the line
perpendicular to the tangent line at the point of tangency. Show that
this line goes through the center of the circle.
Solution:
To check that the line is tangent to the circle, substitute the equation
of the line into the equation for the circle.
x 2  (5  2 x) 2  5
x 2  25  20 x  4 x 2  5
5 x 2  20 x  20  0
x 2  4x  4  0
( x  2) 2  0
This has a double root at x = 2. This means that the line intersects the circle at only one point (2, 1).
A perpendicular line to the tangent line would have slope that is the negative reciprocal of the tangent line
or m 
1
.
2
The equation of the line can be written: y 
1
xb .
2
We find the value of b by plugging in the tangency point: (2, 1)
1
1
(2)  b  b  0
2
The equation is y 
1
x and we know that it passes through the origin since the y-intercept is zero.
2
This means that the radius of the circle is perpendicular to the tangent to the circle.
Homework:
Determine whether each segment is tangent to the given circle:
1.
2.
3.
O
O
18
7
8
6
O
10
12
26
24
24
Find the measure of angle x.
4.
5.
6.
O
O
65 
x
50 
O
x
x
Find the missing length:
7.
8.
9.
20
9
x
O
12
6
x
x
O
8
O
x
4
Find the values of the missing variables
10.
11.
12.
8
O
O
x
z
x
8
5
12
5
12
8
10
13. Find the perimeter of the pentagon:
A
12 B
14. Find the perimeter of the parallelogram:
C
15
7
21
D
J
E
8
I
F
5
H
24
G
15. Find the perimeter of the right triangle:
16. Find the perimeter of the polygon:
10
12
15
4
5
6
Determine if the circles given by the equations are concentric.
17. x 2  4 x  y 2  2 y  4  0 and x 2  4x  y 2  2 y  31  0
18. x 2  6x  y 2  8 y  0 and x 2  6x  y 2  8 y  9  0
19. x 2  y 2  21  10 y and x 2  y 2  8 y  33
20. x 2  y 2  26  10x  10 y and x 2  y 2  10x  10 y  14
21. Draw the line y = 3x + 10 and the circle x2 + y2 = 10. Show that these graphs touch at only one point.
Find the slope of the segment that joins this point to the center of the circle, and compare your answer with
the slope of the line y = 3x +10.
Solutions:
1. yes
2. yes
3. no
4. 25o
5. 10o
6. 72o
7. 12.6
8. 10.67
9. 8.1
10. 3.9
11. 9.6
12. 12.8
13. 128
14. 112
15. 9.3
16. 76
17. yes
18. no
19. no
20. yes
21. x2 + (3x+10)2 = 10 solve for x to obtain double root (-3, 1).
Slope of line from (0, 0) to (-3, 1) = -1/3 which is negative reciprocal of slope of the line.
This means that tangent line and radius are perpendicular.