What is (x-3)^2+y^2=8y-12 ?

The solution to (x-3)^2+y^2=8y-12 is Circle with (a,b)=(3,4),r=2

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(x-3)^2+y^2=8y-12

(x - 3)^{2} + y^{2} = 8 y - 12

(a, b) = (3, 4), r = 2

Solution steps

(x - 3)^{2} + y^{2} = 8 y - 12

Rewrite

(x - 3)^{2} + y^{2} = 8 y - 12

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, 4), r = 2

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

y^{2} = 9 - x^{2}

x^{2} + y^{2} - 4 x - 6 y - 23 = 0

y^{2} + z^{2} = 9

(x - 3)^{2} + (y - 5)^{2} = 121

What is (x-3)^2+y^2=8y-12 ?
The solution to (x-3)^2+y^2=8y-12 is Circle with (a,b)=(3,4),r=2