Question 1086213
The center of the circle lies on the line perpendicular to the tangent line and passing through (3,-3).
Putting x-4y-15=0 in standard form we have y = x/4 - 15/4
Thus the equation of the line passing through the center is:
y + 3 = -4(x - 3) -> y = -4x + 9
The distance from the center of the circle to each of the two points is equal to the radius of the circle, by definition.
Let the center of the circle be the point (a,b)
Therefore sqrt((a-3)^2 + (b+3)^2) = sqrt((a-6)^2 + (b-2)^2)
And since the center lies on y = -4x + 9, we can substitute for b:
b = -4a + 9
Making the substitution, squaring both sides and collecting terms we are left with:
34a = 68 or a = 2
Therefore b = -4*2 + 9 = 1
So the center is (2,1) and the radius R = sqrt((2-3)^2+(1+3)^2) = sqrt(17)
Thus the equation of the circle is (x-2)^2 + (y-1)^2 = 17