Find the center of a circle with the equation: x2+y2-32x-60y+1122=0

(16, 30)

Step-by-step explanation:

First the equation of a circle is:

(x-h) ^2 + (y-k) ^2 = r^2, where (h, k) - the center.

We rewrite the equation and set them equal:

(x-h) ^2 + (y-k) ^2 - r^2 = x^2+y^2 - 32x - 60y + 1122=0

x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y + 1122 = 0

We solve for each coeffiecient meaning if the term on the LHS contains an x then its coefficient is the same as the one on the RHS containing the x or y.

-2hx = - 32x = > h = 16.

-2ky = - 60y = > k = 30. = > the center is at (16, 30)