Geometry H
Section 10/5 – 10/7
Review Worksheet answers
For exercises # 1 – 3
1. mRS = 80. mUT = 110.
m1 = ___
m1 = ½ (mRS + mUT)
= ½ (80 + 110)
= ½ (190) = 95
2. QR = 8. QT = 9. QU = 2QS.
QS = ___
QUQS = QRQT
2QUQU = 89
2(QU)2 = 72
(QU)2 = 36
QU = 6
3. QU = 4. QS = 6. RT = 11.
QR > QT.
QT = ___
QRQT = QUQS
Since QR + QT = RT, QR = RT - QT
(RT – QT)QT = 46
(11 – QT)QT = 24
11QT – (QT)2 = 24
0 = (QT)2 – 11QT + 24
0 = (QT – 3)(QT – 8)
QT – 3 = 0 or QT – 8 = 0
QT = 3 or QT = 8
If QT = 3, then QR = 11 – 3 = 8
If QT = 8, then QR = 11 – 8 = 3
Since QR > QT, QT = 3
For exercises # 4 – 6
4. mDF = 80. mX = 20.
mCE = ____
mX = ½ (mDF – mCE)
20 = ½ (80 - mCE)
40 = 80 - mCE
-40 = - mCE
40 = mCE
5. XC = 6. CD = 4. XE = 5.
EF = ___
XEXF = XGXD
XE(XE + EF) = XC(XC + CD)
5(5 + EF) = 6(6 + 4)
25 + 5EF = 60
5EF = 35
EF = 7
6. XE = 4. EF = 8. CD = 13.
XC = ___
XC∙XD =
XE∙XF
XC(XC + CD) = XE(XE + EF)
XC(XC + 13) = 4(4 + 8)
(XC)2 + 13∙XC = 48
(XC)2 + 13∙XC - 48 = 0
(XC – 3)(XC + 16) = 0
XC – 3 = 0 or XC + 16 = 0
XC = 3 or XC = -16
Since XC > 0, XC = 3
Geometry H
Section 10/5 – 10/7
Review Worksheet answers
For exercises # 7 – 9
Give the center and the radius of the circle.
11. (x – 3)2 + (y+9)2 = 25
Since (x-h)2 + (y-k)2 = r2,
(x – 3)2 + (y+9)2 = 25
This can be expressed as
(x – 3)2 + (y- -9)2 = 52
Center : (3,-9) r : 5
YZ is a tangent ray.
7. mXK = 40. mXYZ = 50.
mKY = ___
Since XYZ is a secant-tangent angle,
mXYZ = ½ mXKY.
By Arc Addition, mXKY = mXK + mKY.
So, mXYZ = ½ (mXK + mKY).
50 = ½ (40 + mKY)
100 = 40 + mKY
60 = mKY
8. mXJY = 5mXKY. mXYZ = ___
By definition of the measure of a major arc,
mXJY = 360 - mXKY
5mXKY = 360 - mXKY
6mXKY = 360
mXKY = 60
Since XYZ is a secant-tangent angle,
mXYZ = ½ mXKY.
= ½  60 = 30
9.
JX  XY  YJ
mXPY = ___
Since JX  XY  YJ ,
mJX = mXY = mYJ
mXY = 360  3 = 120
Since XPY is a central angle,
mXPY = mXY = 120
12. x2 + (y+2)2 = 8
Since (x-h)2 + (y-k)2 = r2,
x2 + (y+2)2 = 8
This can be expressed as
(x – 0)2 + (y- -2)2 = (2 2 )2
Center : (0,-2) r : 2 2
Give the equation in center-radius form of
the circle.
13. with center (-4, 3) and radius 7
Since (x-h)2 + (y-k)2 = r2,
(x – -4)2 + (y – 3)2 = 72
(x + 4)2 + (y – 3)2 = 49
14. with center (1,0) and containing (2,3)
Since (x-h)2 + (y-k)2 = r2,
(x – 1)2 + (y – 0)2 = r2
(x - 1)2 + y2 = r2
When x = 2, y = 3
(2 - 1)2 + 32 = r2
1
+ 9 = r2
r2 = 10
(x - 1)2 + y2 = 10
15. with center (3,7) and tangent to the y-axis.
The distance from (3,7) to the y-axis is 3.
So, r = 3.
Since (x-h)2 + (y-k)2 = r2,
(x - 3)2 + (y – 7)2 = 32
(x - 3)2 + (y – 7)2 = 9
Geometry H
Section 10/5 – 10/7
Review Worksheet answers
16. with center (1,1) and tangent to
the line y = 5.
The distance from (1,1) to the line y = 5 is 4.
So, r = 4.
Since (x-h)2 + (y-k)2 = r2,
(x - 1)2 + (y – 1)2 = 42
(x - 1)2 + (y – 1)2 = 16
17. with a diameter whose endpoints are
(5,9) and (11,17)
The center of a circle lies at the midpoint of a
diameter.
Using the midpoint formula,
h = (5+11)/2 = 16/2 = 8
k = (9+17)/2 = 26/2 = 13
The length of the radius is half the length of a
diameter.
Using the distance formula,
r = ½ (11  5) 2  (17  9) 2
= ½ 36  64
= ½ 100 = ½(10) = 5
Since (x-h)2 + (y-k)2 = r2,
(x – 8)2 + (y – 13)2 = 52
(x – 8)2 + (y – 13)2 = 25
18. with the equation
x2 + y2 – 6x + 2y + 1 = 0
x2 + y2 – 6x + 2y = -1
(x2 - 6x ) + (y2 + 2y ) = -1
(x2 - 6x + 32) + (y2 + 2y + 12) = -1+ 32 + 12
(x - 3)2 + (y + 1)2 = 9
19. with the equation
x2 + y2 – 5y + 6 = 0
x2 + y2 – 5y
= -6
2
2
x + (y – 5y ) = -6
x2 + [y2 – 5y + (5/2)2 ] = (5/2)2 + -6
x2 + (y – 5/2)2 = 25/4 - 24/4
x2 + (y – 2.5)2 = 1/4
Put the circle in standard form.
Then find the center and the radius.
20. x2 + y2 + 8x + 6y - 24 = 0
x2 + 8x
+ y2 + 6y
= 24
2
(x + 8x + 42) + (y2 + 6y + 32) = 24 + 42 + 32
(x + 4)2 + (y + 3)2 = 24 + 16 + 9
(x + 4)2 + (y + 3)2 = 49
Center : (-4, -3)
radius : 7
21. x2 + y2 + 3x - 5y + 8 = 0
x2 + 3x
+ y2 - 5y
= -8
x2 + 3x +
 32 2 + y
2
- 5y +
 52 2 = -8 +  32 2 +  52 2
( x  32 ) 2  ( y  52 ) 2 =
32
4
2
2
4
 94  254
(x + 1.5)2 + (y – 2.5) =
Center : (-1.5, 2.5)
radius : 22
Give the equation of the circle containing
the three points. Then find the center and the
radius.
22. (-4,2) (-6,-2) (-1,3)
Since x2 + y2 + Dy + Ey + F = 0
For (-4,2), 16 + 4 + -4D + 2E + F = 0
For (-6,-2), 36 + 4 + -6D + -2E + F = 0
For (0,3),
1 + 9 + -1D + 3E + F = 0
So,
-4D + 2E + F = -20
-6D + -2E + F = -40
-1D + 3E + F = -10
Subtracting Equation 2 from Equation 1 and
subtracting Equation 3 from Equation 2,
2D + 4E = 20
-5D + -5E = -30
Dividing the top equation by 2
and the bottom one by 5,
Geometry H
Section 10/5 – 10/7
Review Worksheet answers
D + 2E = 10
-D + -E = -6
Adding the two equations,
E = 4
If E = 4, D + 8 = 10. D = 2
Substituting into Equation 1,
-4(2) + 2(4) + F = -20
F = -20
x2 + y2 +2x + 4y – 20 = 0
x2 + y2 +2x + 4y
= 20
(x2 + 2x ) + (y2 + 4y ) = 20
(x2 + 2x + 12) + (y2 + 4y + 22) = 20 + 12 + 22
(x + 1)2 + (y + 2)2 = 25
Center : (-1, -2)
radius : 5
23. (0,12) (-10, 2) (8,8)
Since x2 + y2 + Dy + Ey + F = 0
For (0,12),
0 + 144 + 0D + 12E + F = 0
For (-10, 2), 100 + 4 + -10D + 2E + F = 0
For (8,8),
64 + 64 + 8D + 8E + F = 0
So,
0D + 12E + F = -144
-10D + 2E + F = -104
8D + 8E + F = -128
Subtracting Equation 2 from Equation 1 and
subtracting Equation 3 from Equation 2,
10D + 10E = -40
-18D + -6E = 24
Dividing the top equation by 10
and the bottom one by 6,
D + E = -4
-3D + -E = 4
Adding the equations,
-2D
= 0
So, D = 0
If D = 0, E = -4
Substituting into Equation 1,
0(0) + 12(-4) + F = -144
-48 + F = -144
F = -96
x2 + y2 + 0x - 4y – 96 = 0
x2 + y2 - 4y – 96 = 0
x2 + y2 - 4y
= 96
x2 + (y2 - 4y ) = 96
x2 + (y2 - 4y + 42) = 96 + 42
x2 + (y – 4)2 = 112
Center : (0, 4)
radius : 4 7