What is x^2+y^2-32x-8y+264=0 ?

The solution to x^2+y^2-32x-8y+264=0 is Circle with (a,b)=(16,4),r=2sqrt(2)

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x^2+y^2-32x-8y+264=0

x^{2} + y^{2} - 32 x - 8 y + 264 = 0

(a, b) = (16, 4), r = 22

Solution steps

x^{2} + y^{2} - 32 x - 8 y + 264 = 0

Rewrite

x^{2} + y^{2} - 32 x - 8 y + 264 = 0

in the form of the standard circle equation

(x - 16)^{2} + (y - 4)^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (16, 4), r = 22

(x - 2)^{2} + (y + 5)^{2} = 49

\frac{( x + 5 ) ^{2}}{\frac{9}{25}} + \frac{( y + 4 ) ^{2}}{\frac{9}{25}} = 1

x^{2} + y^{2} - 8 x + 14 y + 16 = 0

(x - 0)^{2} + (y - 0)^{2} = 9

(x + 3)^{2} + (y + 4)^{2} = 9

What is x^2+y^2-32x-8y+264=0 ?
The solution to x^2+y^2-32x-8y+264=0 is Circle with (a,b)=(16,4),r=2sqrt(2)