What is 2x^2+8x+2y^2+4y-62=0 ?

The solution to 2x^2+8x+2y^2+4y-62=0 is Circle with (a,b)=(-2,-1),r=6

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2x^2+8x+2y^2+4y-62=0

2 x^{2} + 8 x + 2 y^{2} + 4 y - 62 = 0

(a, b) = (- 2, - 1), r = 6

Solution steps

2 x^{2} + 8 x + 2 y^{2} + 4 y - 62 = 0

Rewrite

2 x^{2} + 8 x + 2 y^{2} + 4 y - 62 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 1))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 1), r = 6

x^{2} + y^{2} + 4 x + 6 y + 4 + 9 = 12

x^{2} + y^{2} - 20 y + 36 = 0

3 (x + \frac{3}{4})^{2} + 3 (y - \frac{1}{2})^{2} = 36

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

(x - 10)^{2} + (y + 8)^{2} = 36

What is 2x^2+8x+2y^2+4y-62=0 ?
The solution to 2x^2+8x+2y^2+4y-62=0 is Circle with (a,b)=(-2,-1),r=6