What is x^2+y^2-8y+12=0 ?

The solution to x^2+y^2-8y+12=0 is Circle with (a,b)=(0,4),r=2

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x^2+y^2-8y+12=0

x^{2} + y^{2} - 8 y + 12 = 0

(a, b) = (0, 4), r = 2

Solution steps

x^{2} + y^{2} - 8 y + 12 = 0

Rewrite

x^{2} + y^{2} - 8 y + 12 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 4)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (0, 4), r = 2

(x - 6)^{2} + (y - 10)^{2} = (5)^{2}

(x + 6)^{2} + (y + 8)^{2} = 100

3 x^{2} + 3 y^{2} - 12 x - 24 y + 12 = 0

(x - 5)^{2} + (y - 7)^{2} = 4

x^{2} + y^{2} - 4 x + 12 y - 9 = 0

What is x^2+y^2-8y+12=0 ?
The solution to x^2+y^2-8y+12=0 is Circle with (a,b)=(0,4),r=2