Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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

Sagot :

The equation of a circle exists:

[tex]$(x-h)^2 + (y-k)^2 = r^2[/tex], 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:

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

We rewrite the equation and set them equal :

[tex]$(x-h)^2 + (y-k)^2 - r^2 = x^2+y^2- 32x - 60y +1122=0[/tex]

[tex]$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y +1122 = 0[/tex]

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:

https://brainly.com/question/10633821

#SPJ4