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)

1,589.63 in.^3

step-by-step explanation:

i hope this you : d