A circle with centre (5,-4) passes through the point (5, 0). Find its equation.
x\(^2\) + y\(^2\) + 10x + 8y + 25 =0
x\(^2\) + y\(^2\) +10x - 8y - 25 = 0
x\(^2\) + y\(^2\) - 10x + 8y + 25 =0
x\(^2\) + y\(^2\) -10x - 8y - 25 = 0
Correct answer is C
(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)
x\(^2\) - 2hx + y\(^2\) - 2ky + h\(^2\) + k\(^2\) = r\(^2\)
x\(^2\) - 2(3)x + y\(^2\) - 2(-4) y + 5\(^2\) + (-4)\(^2\) = r\(^2\)
x\(^2\) - 10x + y\(^2\) + 8y + 25 + 16 = r\(^2\)
x\(^2\) - 10x + y\(^2\) + 8y + 41 = r\(^2\)
at point (5,0)
5\(^2\) - 10(5) + 0\(^2\) + 8(0) + 41 = r\(^2\)
25 - 50 + 41 = r\(^2\)
16 = r\(^2\)
r = \(\sqrt{16}\)
= 4
x\(^2\) + y\(^2\) - 10x + 8y + 25 = 0