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!!

Noori Aug 9, 2020

#1**+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!

OlympusHero Aug 9, 2020

#3**+1 **

For the first one:

1) Extend QRS to point T on the circle.

2) By the "power of a point": QP^{2} = QR·QT ---> 6^{2} = 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.

OT^{2} = r^{2} = 6^{2} + 2^{2} - 2·6·2·cos(TSO)

OR^{2} = r^{2} = 3^{2} + 2^{2} - 2·3·2·cos(RSO)

Since these are equal: 6^{2} + 2^{2} - 2·6·2·cos(TSO) = 3^{2} + 2^{2} - 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 OT^{2} = r^{2} = 6^{2} + 2^{2} - 2·6·2·cos(TSO)

---> r^{2} = 40 - 24·0.75

r^{2} = 22

r = sqrt(22)

geno3141 Aug 9, 2020

#4**+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

Dragan Aug 10, 2020

#6**+3 **

1) Extend line segment QS to point T on a circle.

QT = PQ^{2} / RQ RT = QT - RQ

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

MO = sqrt( OS^{2} - MS^{2 })

Radius OR = sqrt( MR^{2} + MO^{2} )

Dragan
Aug 10, 2020