+82 6)Two circles intersect at A and B. A common external tangent is tangent to the circles at T and U as shown. Let M be the intersection of line AB and TU. If AB = 9 and BM = 3 find TU.
Picture: https://latex.artofproblemsolving.com/c/5/0/c50a548001dc09c6eb63031398f6a77787276a50.png
8)Circle O is tangent to AB at A, and angle ABD = 90 degrees. If AB = 12 and CD = 18, find the radius of the circle.
Picture: https://latex.artofproblemsolving.com/f/f/1/ff13ec3e917b2a46649871ef2354fc1290478bc0.png
9)In cyclic quadrilateral ABCD, AB = 2, BC = 3, CD = 10, and DA = 6. Let P be the intersection of lines AB and CD. Find the length BP.
Picture: https://latex.artofproblemsolving.com/d/a/2/da27e9a157dbaae3feaf8dce05d22b710c368627.png
10)Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB. Let P be a point on OA, and let line CP intersect the circle again at Q. If OP = 20 and PQ = 7, find r^2, where r is the radius of the circle.
Picture: https://latex.artofproblemsolving.com/5/2/0/52028bb1d6c8a816ada5d2be3e55df12d704da4a.png
11)Two parallel chords in a circle have lengths 18 and 14, and the distance between them is 8. Find the length of the chord parallel to these chords and midway between them. (The diagram is not drawn to scale.)https://latex.artofproblemsolving.com/c/0/6/c06a5e099c8cb008ed2811aa67bc29aa48e94b8a.png
+129907 8)Circle O is tangent to AB at A, and angle ABD = 90 degrees. If AB = 12 and CD = 18, find the radius of the circle.
We have the secant-tangent theorem, first
AB^2 = CB (CB + CD)
12^2 = CB^2 + 18CB
CB^2 + 18CB - 144 = 0 factor as
(CB - 6) ( CB + 18) = 0
The frist answer gives the positive answer for CB - 6 = 0 ⇒ CB = 6
Draw AC ......and the tangent of angle BAC = 6/12 = 1/2
So arctan (1/2) = BAC....and minor arc AC is twice this = 2 arctan (1/2)
And AC = √ ( BA^2 + BC^2) = √ (12^2 + 6^2) = √ (144 + 36) = √160
Draw OC and OA and angle COA = minor arc AC = 2 arctan (1/2)
Using the Law of Cosines to find the radius.....we have that
160 = 2r^2 - 2r^2 * cos (2 arctan (1/2) )
Let cos (2arctan(1/2) ) = 1 - 2sin^2 ( arctan(1/2) ) ⇒ sin (arctan (1/2) ) = 1/ √5 ....so we have
1 - 2(1/√5)^2 = 1 - 2/5 = 3/5 = .6
160 = 2r^2 - 2r^2 (.6)
160 = r^2 ( 2 - 1.2)
160 = r^2 (.8)
200 = r^2
r = √200 = 10√2
+129907 9)In cyclic quadrilateral ABCD, AB = 2, BC = 3, CD = 10, and DA = 6. Let P be the intersection of lines AB and CD. Find the length BP.
Draw AC
AC^2 = 2^2 + 3^2 - 2(3)(2)cos (ABC)
AC^2 = 13 - 12cos(ABC) (1)
And since ABC is obtuse and ADC is supplemental to ABC, cos(ADC) = -cos(ABC)
So
AC^2 = 6^2 + 10^2 - 2(6)(10)cos(ADC)
AC^2 = 6^2 + 10^2 - 120 [ -cos(ABC) ]
AC^2 = 6^2 + 10^2 + 120cos(ABC)
AC^2 = 136 + 120cos(ABC) (2)
Subtract (1) from (2)
0 = 123 + (120 + 12) cos(ABC)
0 = 123 + 132cos(ABC)
-123 / 132 = cos (ABC)
-41/44 = cos(ABC)
So sin (ABC) = √ [ 1 - (41/44)^2 ] = √ (255) / 44 = sin PBC
Likewise.....draw DB
DB^2 = 6^2 + 2^2 - 2(6)(2)cos(DAB)
DB^2 = 40 - 24cos(DAB)
And sincw DAB is obtuse and DCB is supplemental to DCB, cos(DCB) = -cos(DAB)
So
DB^2 = 10^2 + 3^2 - 2(10)(3)cos(DCB)
DB^2 = 10^2 + 3^2 - 60 [-cos(DAB) ]
DB^2 = 109 + 60cos(DAB)
Subtract (3) from (4)
0 = 69 + (60 + 24) cos(DAB)
-69/84 = cos(DAB)
-23/28 = cos(DAB)
So sin (DAB) = √ [ 1 - (23/28)^2 ] = √ (255)/28 = sin PCB
sin PBC / sin PCB = PC/ PB
(√ (255) / 44) / ( √ (255)/28) = PC / PB
28 / 44 = PC /PB
7/11 = PC /PB
PC = (7/11)PB
And we have that
(PB + AB) * PB = PC(PC + CD)
(PB + 2) * PB = (7/11)PB [ (7/11)PB + 10)
PB^2 + 2PB = (49/121)PB^2 + (70/11)PB
Let PB = x
x^2 + 2x = (49/121)x^2 + (70/11)x
(121 - 49) x^2 / 121 + (2 - 70/11)x = 0
(72/121)x^2 - (48/11)x = 0
(1/11)x [ (72/11)x - 48 ] = 0
So..... either x = 0 {reject} or
(72/11)x - 48 = 0
(72/11)x = 48
x = 48(11)/72 = (2/3) * 11 = 22/3 = PB = BP
Proof {PC = (7/11)PB = (7/11)(22/3) = 14/3 }
(PB + AB) * PB = PC(PC + CD)
(22/3 + 2) * (22/3) = (14/3) * ( 14/3 + 10)
616 / 9 = 616 / 9
