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.
+128408 See the image below :
Let A =(8,0)
Let B = (0,6)
Let C = (0,0)
The center of the circumcenter lies at the midpoint of the hypotenuse = (4,3)
And the radius of the ctrcumcircle = 5
So.....the equation of the circumcircle =
(x - 4)^2 + (y-3)^2 = 25 (1)
And since angle ACB = 90° then the angle bisector will be a 45° line through the origin
And the equation of this line is y = x
Substiting this into (1) for x gives the x coordinate of M
So
(x - 4)^2 + ( x -3)^2 = 25
x^2 -8x + 16 + x^2 -6x + 9 = 25 simplify
2x^2 - 14x = 0
2x ( x - 7) = 0
Setting each factor to 0 and solving for x produces x = 0 or x = 7
We want the second result
So the y coordinate of M = 7
So....the length of CM = sqrt (7^2 + 7^2) = 7sqrt(2) units
