What is 2x^2+2y^2-12x+4y-15=0 ?

The solution to 2x^2+2y^2-12x+4y-15=0 is Circle with (a,b)=(3,-1),r=sqrt(35/2)

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2x^2+2y^2-12x+4y-15=0

2 x^{2} + 2 y^{2} - 12 x + 4 y - 15 = 0

(a, b) = (3, - 1), r = \frac{35}{2}

Solution steps

2 x^{2} + 2 y^{2} - 12 x + 4 y - 15 = 0

Rewrite

2 x^{2} + 2 y^{2} - 12 x + 4 y - 15 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 1))^{2} = (\frac{35}{2})^{2}

Therefore the circle properties are:

(a, b) = (3, - 1), r = \frac{35}{2}

(x - 2)^{2} + y^{2} = 1

x^{2} + y^{2} - 2 x = 3

x^{2} + y^{2} - 12 x + 8 y - 5 = 0

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

16 x^{2} + 16 y^{2} + 96 x - 40 y - 315 = 0

What is 2x^2+2y^2-12x+4y-15=0 ?
The solution to 2x^2+2y^2-12x+4y-15=0 is Circle with (a,b)=(3,-1),r=sqrt(35/2)