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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
avatar+75 
+1

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
avatar+215 
-2

Ok, thank you! I will try this out. 

Noori  Aug 9, 2020
 #5
avatar+215 
-2

Can you help me with the steps, for the second one? I don't get how it works. 

Noori  Aug 10, 2020
 #3
avatar+21955 
+1

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).

      OT and OR are radii; Let the radius = r.

 

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
avatar+1361 
+3

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
avatar+1361 
+3

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
avatar+215 
-1

Can I have help with the steps for the second one? Could someone explain it?

Noori  Aug 10, 2020
 #8
avatar+215 
-1

Hi, I don't know trig yet. Is there a different way of explaining the steps?

Noori  Aug 10, 2020
 #9
avatar+111089 
0

Noori,

Please post just one question per post.

 Aug 11, 2020

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