1. Warm Up
Write the equation of each item.
1. FG x = –2
y = 32. EH
3. 2(25 –x) = x + 2 4. 3x + 8 = 4x
x = 16 x = 8
GT Geometry Drill 5/27/14

2. Identify tangents, secants, and
chords.
Use properties of tangents to solve
problems.
Objectives

3. interior of a circle concentric circles
exterior of a circle tangent circles
chord common tangent
secant
tangent of a circle
point of tangency
congruent circles
Vocabulary

4. The interior of a circle is the set of all
points inside the circle. The exterior of a
circle is the set of all points outside the
circle.

6. Check It Out! Example 1
Identify each line or segment that
intersects P.
chords:
secant:
tangent:
diameter:
radii:
QR and ST
ST
PQ, PT, and PS
UV
ST

8. Find the length of each radius. Identify the point
of tangency and write the equation of the tangent
line at this point.
Example 2: Identifying Tangents of Circles
radius of R: 2
Center is (–2, –2). Point on
 is (–2,0). Distance
between the 2 points is 2.
Center is (–2, 1.5). Point on 
is (–2,0). Distance between
the 2 points is 1.5.
radius of S: 1.5

9. Find the length of each radius. Identify the point
of tangency and write the equation of the tangent
line at this point.
Example 2 Continued
point of tangency: (–2, 0)
Point where the s and
tangent line intersect
equation of tangent line: y = 0
Horizontal line through (–2,0)

10. Check It Out! Example 2
Find the length of each radius. Identify the point
of tangency and write the equation of the tangent
line at this point.
radius of C: 1
Center is (2, –2). Point on 
is (2, –1). Distance between
the 2 points is 1.
radius of D: 3
Center is (2, 2). Point on  is
(2, –1). Distance between the
2 points is 3.

11. Check It Out! Example 2 Continued
Find the length of each radius. Identify the point
of tangency and write the equation of the tangent
line at this point.
Pt. of tangency: (2, –1)
Point where the s and
tangent line intersect
eqn. of tangent line: y = –1
Horizontal line through (2,-1)

12. A common tangent is a line that is tangent to
two circles.

13. A common tangent is a line that is tangent to
two circles.

15. Example 3: Problem Solving Application
Early in its flight, the Apollo 11 spacecraft
orbited Earth at an altitude of 120 miles.
What was the distance from the
spacecraft to Earth’s horizon rounded to
the nearest mile? The radius of the earth
is 4000 miles
The answer will be the length of an
imaginary segment from the spacecraft
to Earth’s horizon.
1 Understand the Problem

16. 2 Make a Plan
Draw a sketch. Let C be the
center of Earth, E be the
spacecraft, and H be a point
on the horizon. You need to
find the length of EH, which
is tangent to C at H. By
Theorem 11-1-1, EH  CH.
So ∆CHE is a right triangle.

17. Solve3
EC = CD + ED
= 4000 + 120 = 4120 mi
EC2 = EH² + CH2
41202 = EH2 + 40002
974,400 = EH2
987 mi  EH
Seg. Add. Post.
Substitute 4000 for CD
and 120 for ED.
Pyth. Thm.
Substitute the
given values.
Subtract 40002 from
both sides.
Take the square root of
both sides.

18. Look Back4
The problem asks for the distance to the
nearest mile. Check if your answer is
reasonable by using the Pythagorean
Theorem. Is 9872 + 40002  41202?
Yes, 16,974,169  16,974,400.

19. Check It Out! Example 3
Kilimanjaro, the tallest mountain in
Africa, is 19,340 ft tall. What is the
distance from the summit of Kilimanjaro
to the horizon to the nearest mile?
The answer will be the length of an
imaginary segment from the summit of
Kilimanjaro to the Earth’s horizon.
1 Understand the Problem

20. 2 Make a Plan
Draw a sketch. Let C be the
center of Earth, E be the
summit of Kilimanjaro, and H
be a point on the horizon. You
need to find the length of EH,
which is tangent to C at H. By
Theorem 11-1-1, EH  CH.
So ∆CHE is a right triangle.

21. Solve3
EC = CD + ED
= 4000 + 3.66
= 4003.66mi
EC2 = EH2 + CH2
4003.662 = EH2 + 40002
29,293 = EH2
171  EH
Seg. Add. Post.
Substitute 4000 for CD and
3.66 for ED.
Pyth. Thm.
Substitute the given values.
Subtract 40002 from both sides.
Take the square root of
both sides.
ED = 19,340 Given
Change ft to mi.

22. Look Back4
The problem asks for the distance from the
summit of Kilimanjaro to the horizon to the
nearest mile. Check if your answer is
reasonable by using the Pythagorean
Theorem. Is 1712 + 40002  40042?
Yes, 16,029,241  16,032,016.

24. Example 4: Using Properties of Tangents
HK and HG are tangent to F. Find HG.
HK = HG
5a – 32 = 4 + 2a
3a – 32 = 4
2 segments tangent to
 from same ext. point
 segments .
Substitute 5a – 32 for
HK and 4 + 2a for HG.
Subtract 2a from both sides.
3a = 36
a = 12
HG = 4 + 2(12)
= 28
Add 32 to both sides.
Divide both sides by 3.
Substitute 12 for a.
Simplify.

25. Check It Out! Example 4a
RS and RT are tangent to Q. Find RS.
RS = RT
2 segments tangent to 
from same ext. point 
segments .
x = 8.4
x = 4x – 25.2
–3x = –25.2
= 2.1
Substitute 8.4 for x.
Simplify.
x
4
Substitute for RS and
x – 6.3 for RT.
Multiply both sides by 4.
Subtract 4x from both sides.
Divide both sides by –3.

26. Check It Out! Example 4b
n + 3 = 2n – 1 Substitute n + 3 for RS
and 2n – 1 for RT.
4 = n Simplify.
RS and RT are tangent to Q. Find RS.
RS = RT
2 segments tangent to 
from same ext. point 
segments .
RS = 4 + 3
= 7
Substitute 4 for n.
Simplify.

27. Holt McDougal Geometry
11-1 Lines That Intersect Circles
A central angle is an angle whose
vertex is the center of a circle. An arc is
an unbroken part of a circle consisting of
two points called the endpoints and all
the points on the circle between them.

28. Holt McDougal Geometry
11-1 Lines That Intersect Circles

29. Holt McDougal Geometry
11-1 Lines That Intersect Circles

30. Holt McDougal Geometry
11-1 Lines That Intersect Circles

31. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Example 1: Data Application
The circle graph shows the types of grass
planted in the yards of one neighborhood.
Find mKLF.
= 234
mKLF = 360° – mKJF
mKJF = 0.35(360)
= 126
mKLF = 360° – 126°

32. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Adjacent arcs are arcs of the same
circle that intersect at exactly one point.
RS and ST are adjacent arcs.

33. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Example 2: Using the Arc Addition Postulate
mCFD = 180 – (97.4 + 52)
= 30.6
= 97.4 + 30.6
= 128
mBD = mBC + mCD
mBC = 97.4 Vert. s Thm.
∆ Sum Thm.
mCFD = 30.6
Arc Add. Post.
Substitute.
Simplify.
Find mBD.
mCD = 30.6

34. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Check It Out! Example 2a
Find each measure.
mJKL
mKPL = 180° – (40 + 25)°
= 25° + 115°
mKL = 115°
mJKL = mJK + mKL
= 140°
Arc Add. Post.
Substitute.
Simplify.

35. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Check It Out! Example 2b
Find each measure.
mLJN
= 295°
mLJN = 360° – (40 + 25)°

36. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Within a circle or congruent circles,
congruent arcs are two arcs that have the
same measure. In the figure ST  UV.

37. Holt McDougal Geometry
11-1 Lines That Intersect Circles

38. Holt McDougal Geometry
11-1 Lines That Intersect Circles

39. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Find NP.
Example 4: Using Radii and Chords
Step 2 Use the Pythagorean Theorem.
Step 3 Find NP.
RN = 17 Radii of a  are .
SN2 + RS2 = RN2
SN2 + 82 = 172
SN2 = 225
SN = 15
NP = 2(15) = 30
Substitute 8 for RS and 17 for RN.
Subtract 82 from both sides.
Take the square root of both sides.
RM  NP , so RM bisects NP.
Step 1 Draw radius RN.

40. Holt McDougal Geometry
11-1 Lines That Intersect Circles
Check It Out! Example 4
Find QR to the nearest tenth.
Step 2 Use the Pythagorean Theorem.
Step 3 Find QR.
PQ = 20 Radii of a  are .
TQ2 + PT2 = PQ2
TQ2 + 102 = 202
TQ2 = 300
TQ  17.3
QR = 2(17.3) = 34.6
Substitute 10 for PT and 20 for PQ.
Subtract 102 from both sides.
Take the square root of both sides.
PS  QR , so PS bisects QR.
Step 1 Draw radius PQ.

45. Example 1: Identifying Lines and Segments That
Intersect Circles
Identify each line or segment that
intersects L.
chords:
secant:
tangent:
diameter:
radii:
JM and KM
KM
JM
m
LK, LJ, and LM