What is 2x^2+2y^2-8y-24=0 ?

The solution to 2x^2+2y^2-8y-24=0 is Circle with (a,b)=(0,2),r=4

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2x^2+2y^2-8y-24=0

2 x^{2} + 2 y^{2} - 8 y - 24 = 0

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

Solution steps

2 x^{2} + 2 y^{2} - 8 y - 24 = 0

Rewrite

2 x^{2} + 2 y^{2} - 8 y - 24 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 2 y = 15

x^{(2)} + y^{(2)} - 6 x + 8 y + 6 = 0

(x + 3)^{2} + (y - 7)^{2} = 58

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

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

What is 2x^2+2y^2-8y-24=0 ?
The solution to 2x^2+2y^2-8y-24=0 is Circle with (a,b)=(0,2),r=4