The equation of a circle exists:
[tex]$(x-h)^2 + (y-k)^2 = r^2[/tex], where (h, k) be the center.
The center of the circle exists at (16, 30).
Let, the equation of a circle exists:
[tex]$(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:
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