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# Hello, any help is greatly appreciated!

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Hello! Any help with these problems is greatly appreciated!

A circle is centered at O. The tangent to the circle at P is extended to Q. Line segment QS intersects the circle at R Given that. OS = 2, SR = RQ = 3 and PQ = 6, find the radius of the circle.

Let AB be a diameter of a circle, and let C be a point on the circle such that AC=8 and BC=6. The angle bisector of angle ACB intersects the circle at point M. Find CM.

Thank you so much!!

Aug 9, 2020

#1
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Hello fellow AoPSer!

For the first one you should extend QS such that it intersects the circle again at T. And extend OS to intersect the circle at U and V. Try to find ST from there and set variables to solve an equation with power of a point.

For the second one. Let the intersection of AB and MC be D. Drop an altitude from D to AC. Then use the Angle Bisector Theorem, Similar Triangles, and lastly Power of a Point to get your answer.

Hope this helped!

Aug 9, 2020
#2
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Ok, thank you! I will try this out.

Noori  Aug 9, 2020
#5
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Can you help me with the steps, for the second one? I don't get how it works.

Noori  Aug 10, 2020
#3
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For the first one:

1)  Extend QRS to point T on the circle.

2)  By the "power of a point":  QP2  =  QR·QT   --->   62  =  3·QT   --->   QT = 12

Since QR = 3, RT = 9.

Since RS = 3, ST = 6.

3)  Draw OT and OR, creating triangle(TSO) and triangle(RSO).

4)  Use the Law of Cosines on these two triangles.

OT2  =  r2  =  62 + 22 - 2·6·2·cos(TSO)

OR2  =  r2  =  32 + 22 - 2·3·2·cos(RSO)

Since these are equal:  62 + 22 - 2·6·2·cos(TSO)  =  32 + 22 - 2·3·2·cos(RSO)

40 - 24·cos(TSO)  =  13 - 12·cos(RSO)

Since angle(RSO) is supplementary to angle(TSO),  cos(RSO) = - cos(TSO)

--->                                       40 - 24·cos(TSO)  =  13 + 12·cos(TSO)

27  =  36·cos(TSO)

0.75  =  cos(TSO)

5)  Since  OT2  =  r2  =  62 + 22 - 2·6·2·cos(TSO)

--->            r2  =  40 - 24·0.75

r2  =  22

r  =  sqrt(22)

Aug 9, 2020
#4
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2)   Let's place a point P on a circle opposite to point M. Now we have a right triangle CMP

∠ABC = arctan( AC / BC )

∠CMP = ∠ABC - ∠BCM

CM = cos(∠CMP) * MP

Aug 10, 2020
edited by Dragan  Aug 10, 2020
edited by Dragan  Aug 10, 2020
#6
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1)  Extend line segment QS to point T on a circle.

QT = PQ2 / RQ              RT = QT - RQ

M is a midpoint of line segment RT. Now we have a right triangle OMS.

MO = sqrt( OS2 - MS)

Radius  OR = sqrt( MR2 + MO2 )

Dragan  Aug 10, 2020
#7
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Can I have help with the steps for the second one? Could someone explain it?

Noori  Aug 10, 2020
#8
+216
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Hi, I don't know trig yet. Is there a different way of explaining the steps?

Noori  Aug 10, 2020
#9
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Noori,

Please post just one question per post.

Aug 11, 2020