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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

FiestyGeco  Mar 6, 2018
 #1
avatar
+1

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

Guest Mar 7, 2018
 #2
avatar+87301 
+1

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

 

 

 

 

coolcoolcool

CPhill  Mar 7, 2018
 #3
avatar+87301 
+1

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  

 

 

cool cool cool

CPhill  Mar 7, 2018
 #4
avatar+87301 
+1

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

 

 

cool cool cool

CPhill  Mar 7, 2018
 #5
avatar+87301 
+1

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

 

 

cool cool cool

CPhill  Mar 7, 2018
 #6
avatar+87301 
+1

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

 

 

cool cool cool

CPhill  Mar 7, 2018
 #7
avatar+87301 
+1

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

 

 

cool cool cool

CPhill  Mar 8, 2018

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