+218 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!!
+119 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!
+23252 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)
+1490 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
+1490 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 - MS2 )
Radius OR = sqrt( MR2 + MO2 )
