What is the center x^2+y^2-32x-60y+1122=0 ?

The center x^2+y^2-32x-60y+1122=0 is (16,30)

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center x^2+y^2-32x-60y+1122=0

center

x^{2} + y^{2} - 32 x - 60 y + 1122 = 0

(16, 30)

Solution steps

x^{2} + y^{2} - 32 x - 60 y + 1122 = 0

Rewrite

x^{2} + y^{2} - 32 x - 60 y + 1122 = 0

in the form of the standard circle equation

(x - 16)^{2} + (y - 30)^{2} = (34)^{2}

Therefore the circle properties are:

(a, b) = (16, 30), r = 34

And the center is:

(16, 30)

x^{2}

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

x^{2} - 4 y = 0

\frac{( x - 4 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{9} = 1

x^{2} = 4 y - 2 y^{2}

What is the center x^2+y^2-32x-60y+1122=0 ?
The center x^2+y^2-32x-60y+1122=0 is (16,30)