Question

A circle C passes through the point (-12,9) and is given by the equation (x-a)^2 + (y-b)^2 = r^2. If the equation of the tangent to the given circle at the point (-4,1) is given by x-y+5=0, find the values of a,b and r.

The answers have to be a=-8, b=5 r=5.657

Answer 1

The center of the circle is at x=a, y = b. It is also on a line that is perpendicular to the tangent line. That line has slope -1 and its equation is
(y-1) = -(x+4)
y = -x -3, so since (a,b) is on that line,
b = -a -3

Another equation that relates a to be is
(-12-a)^2 + (9-b)^2
= (-4-a)^2 + (1-b)^2 = r^2
144 +24a + a^2 +81-18b+b^2 = 16 +8a +a^2 +b^2-2b +1
208 +16a -16 b = 0
a - b +13 = 0

Your two equations for a and b can now be solved to get a = -8; b = 5

Insert those values into the original circle equation using any known point on the circle,to get r
(-4+8)^2 + (1-5)^2 = r^2
r = sqrt (16 + 16) = 5.657

Answer 2

thanks!

Answer 3

I should have written:
"(a,b) is also on a line that is perpendicular to the tangent line and passes through its point of contact with the circle", and

"Another equation that relates a to b is.."

The derivation remains the same