+84 1)Chords WY and XZ of a circle are perpendicular. If XV = 4, WV = 3, and VZ = 9, then find YZ.
Picture: https://latex.artofproblemsolving.com/6/1/7/617805ac8c9a1eda9bbf996f32b9282efbd74701.png
2)Chord BA bisects chord OP at S. If AS = 3 and BS = 4, find OP.
3)Chords PQ and RS of a circle meet at X inside the circle. If RS = 38, PX = 6, and QX = 12, then what is the smallest possible value of RX?
4)Let BC and DE be chords of a circle, which intersect at A, as shown. If AB = 3, BC = 15, and DE = 3, then find AE.
Picture: https://latex.artofproblemsolving.com/b/f/3/bf3a913d7231e67c0610502487e0f4d2bbb4745f.png
5)Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the circle again at point B. If PA = 15 and PY = 9, then what is AB?
Picture: https://latex.artofproblemsolving.com/f/2/9/f2914a30663db28178f1400e8a0e4a809b6fd2e3.png
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
7)If AB = 3, AC = 8 and BS = 21, find the area of the circle.
Picture: https://latex.artofproblemsolving.com/7/a/5/7a599f15c79349567f1b440d1e75d954dc5077a4.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
Hi F
i can only help with 2 and 3. I can't see the diagrams of the others.
using the chord intercept theorem in both you get OP = sqrt48 and RX = 2
regards
+128407 1. Intersecting Chord Theorem :
WV * VY = XV * VZ
3 * VY = 4 * 9
VY = 4 * 9 /3 = 12
Since it appears the VY and VZ are perpendicular....then....by the Pythagorean Theorem
VY^2 + VZ^2 = YZ^2
12^2 + 9^2 = YZ^2
225 = YZ^2 take the square root of both sides
15 = YZ
+128407 2)Chord BA bisects chord OP at S. If AS = 3 and BS = 4, find OP
O
B S A
P
Intersecting Chord Theorem
BS * AS = OS * PS but OS = PS....so
BS * AS = OS * OS
3 * 4 = OS^2
12 = OS^2
√12 = OS = 2√3
But OP = OS + PS = OS + OS = 2OS = 2 * 2√3 = 4√3
+128407 3)Chords PQ and RS of a circle meet at X inside the circle. If RS = 38, PX = 6, and QX = 12, then what is the smallest possible value of RX?
RX * XS = PX * QX
RX * XS = 6 * 12
RX * XS = 72
And RX + XS = 38 ⇒ XS = 38 - RX.....so....
RX * XS = 72
RX * (38 - RX ) = 72 rearrange
RX^2 - 38RX + 72 = 0 Factor
(RX - 36) (RX - 2) = 0
And its clear that the smallest value for RX = 2
+128407 4)Let BC and DE be chords of a circle, which intersect at A, as shown. If AB = 3, BC = 15, and DE = 3, then find AE.
Draw BE and CD
We have two similar triangles
Angle BCD = angle BED and angle A is common to both
Therefore
ΔCAD is similar to Δ EAB ⇒ CA / AD = EA / AB
Let AD = x and we have
18 / x = ( x + 3 ) / 3 cross-multiply
18 * 3 = x ( x + 3)
54 = x^2 + 3x rearrange
x^2 + 3x - 54 = 0 factor
(x + 9) ( x - 6 ) = 0
Setting the second factor = 0 and solving for x produces a positive result for x ⇒ 6 = AD
+128407 5)Point Y is on a circle and point P lies outside the circle such that PY is tangent to the circle. Point A is on the circle such that segment PA meets the circle again at point B. If PA = 15 and PY = 9, then what is AB?
This is known as the secant-tangent theorem.....specifically
PY^2 = AP * PB
9^2 = 15 * PB
81 = 15 * PB
PB = 81/15 = 9/ 5
So.....
PB + AB = PA
9/5 + AB = 15
AB = 15 - 9/5
AB = [ 75 - 9 ] / 5 = 66/5
+128407 7) If AB = 3, AC = 8 and BS = 21, find the area of the circle.
AS * AB = AP * AC
(AB + BS) * AB = (AC + CP) * AC
24 * 3 = (8 + CP) * 8
72 = 64 + 8*CP
8 = 8*CP
PC = 1
AP = PC + AC
AP = 1 + 8 = 9
SP = √ ( AS^2 - AP^2) = √ ( 24^2 - 9^2 ) = √495
Draw SC
So..... SC = √ ( SP^2 + PC^2 ) =√ (495 + 1) = √ 496
And SPC is right....so SC is a diameter.of the circle.....so the radius is √496/2
So.....the area of the circle is
pi * (√496 / 2)^2 = 496 * pi / 4 units^2 = 124 pi units^2 ≈ 389.56 units^2
