These problems were extremely difficult and I could not solve them. Help is appreciated.
1. In the diagram below, \(\overline{PQ}\) is tangent at \(P\) to the circle with center \(O\), point \(S\) is inside the circle, and \(\overline{QS}\) intersects the circle at \(R\). If \(QR = RS = 3\), \(OS = 2\), and \(PQ = 6\), then find the radius of the circle.
(Here's an image)
2. In right triangle \(ABC\), the length of side \(\overline{AC}\) is \(8\), the length of side \(\overline{BC}\) is \(6\), and \(\angle C = 90^\circ.\) The circumcircle of triangle \(ABC\) is drawn. The angle bisector of \(\angle ACB\) meets the circumcircle at point \(M.\) Find the length \(CM.\)
Thanks a lot!
+23252 Problem #2: If you are allowed to use coordinate geometry:
Since triangle(ACB) is a right triangle, the center of the circumcenter is the midpoint of the hypotenuse.
Place the diagram on a coordinate plane with the midpoint of the hypotenuse the origin.
Then point C will be the point (-4, -3).
The equation of the bisector of angle(C) has a slope of 1.
This line has an equation of y + 3 = 1(x + 4) ---> y = x + 1.
The circumcenter has its center at (0, 0) and a radius of 5 ---> x2 + y2 = 25.
Combining these two equations will give us the coordinates of point M.
x2 + y2 = 25 and y = x + 1 ---> x2 + (x + 1)2 = 25
---> x2 + x2 + 2x + 1 = 25
---> 2x2 + 2x - 24 = 0
---> x2 + x - 12 = 0
---> (x - 4)(x + 3) = 0
Either x = 4 or x = -3 (impossible)
Since x = 4, by substitution, x = 3 ---> M = (3, 4)
Using the distance formula to find the distance from C(-4, -3) to M(3, 4),
we get the distance of CM to be: 7·sqrt(2).
+397 For #1, extend QRS across the circle to meet the other side at T and drop a perpendicular from the centre O onto this line meeting it at N.
TQ.RQ = PQ^2, so (TR + 3).3 = 36, so TR = 9.
NS = 9/2 - 3 = 3/2.
From the triangle ONS, use Pythagoras to calculate ON^2, and then from the triangle ONT use Pythagoras again to calculate OT, the radius of the circle.
There's a trig method for #2.
Join M to B and A.
Since angle C is a right angle, BA, (length 10) will be a diameter of the circle and in which case angle BMA is also a right angle.
Angles ABM and BAM are both equal to 45 deg, (angles on the same arc), so BM = AM.
Now use Pythagoras in the triangle ABM, BM^2 + AM^2 = 2BM^2 = 10^2 = 100, BM = sqrt(50) = 5sqrt(2).
Call angle CBA theta, then using the sine rule in triangle CBM,
5sqrt(2)/ sin(45) = CM/ sin(theta + 45),
so CM = 5sqrt(2)(sin(theta).cos(45) + cos(theta).sin(45))/sin(45).
Simplify that to get 7sqrt(2), (same as geno).
