Answer:
(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)
The centre of a circle is (16, 30).
The given circle equation is x²+y²−32x−60y+1122=0.
We need to find the centre of a circle for the given equation.
The standard form of the equation is (x-h)² + (y-k)² = r², where (h,k) is the center.
We rewrite the equation and set them equal:
(x-h)² + (y-k)² - r² = 0
x² +y² − 32x − 60y +1122=0
x² - 2hx + h^2 + y² - 2ky + k² - r² = x² + y² - 32x - 60y +1122 = 0
We solve for each coefficient 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.
Therefore, the centre of a circle is (16, 30).
To learn more about the equation of a circle visit:
https://brainly.com/question/10165274.
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