Question

Find the center of a circle with the equation: x2 y2−32x−60y 1122=0 x 2 y 2 − 32 x − 60 y 1122 = 0

Answer 1

The equation of a circle exists:

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

The center of the circle exists at (16, 30).

What is the equation of a circle?

Let, the equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2$, where (h, k) be 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 coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.

-2hx = -32x

h = -32/-2

⇒ h = 16.

-2ky = -60y

k = -60/-2

⇒ k = 30.

The center of the circle exists at (16, 30).

To learn more about center of the circle refer to:

brainly.com/question/10633821

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