What is 2x^2+8x+2y^2-16y-5=0 ?

The solution to 2x^2+8x+2y^2-16y-5=0 is Circle with (a,b)=(-2,4),r=(3sqrt(5))/(sqrt(2))

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2x^2+8x+2y^2-16y-5=0

2 x^{2} + 8 x + 2 y^{2} - 16 y - 5 = 0

(a, b) = (- 2, 4), r = \frac{3 5}{2}

Solution steps

2 x^{2} + 8 x + 2 y^{2} - 16 y - 5 = 0

Rewrite

2 x^{2} + 8 x + 2 y^{2} - 16 y - 5 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 4)^{2} = (\frac{3 5}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, 4), r = \frac{3 5}{2}

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

4 x^{2} + 4 y^{2} - 4 x + 2 y + 1 = 0

(x + 3)^{2} + (y - 1)^{2} = 5

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

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

What is 2x^2+8x+2y^2-16y-5=0 ?
The solution to 2x^2+8x+2y^2-16y-5=0 is Circle with (a,b)=(-2,4),r=(3sqrt(5))/(sqrt(2))