Let A(-8,4) and B(6,6). Point C(0,k) is such that the circumcircle of triangle ABC has equation x^2 + y^2 + 2x - 10y - 24 = 0. Find all possible values of k.
+128474 We can find the center and radius of the circle by completing the square on x and y
x^2 + 2x + 1 + y^2 -10y + 25 = 24 + 1 + 25
(x + 1)^2 + (y - 5)^2 = 50
The center is ( -1, 5) and the radius is sqrt (50)
And we can find the possible values for k by solving this :
(-1 -0)^2 + ( 5 - k)^2 = 50
1 + k^2 - 10k + 25 = 50
k^2 - 10k + 26 = 50
k^2 - 10k - 24 = 0 factor
(k - 12) (k + 2) = 0
Set both factors to 0 and solve for k and we get that
k = 12 or k = -2
See the image here :
