What is 2x^2+2y^2-16x+4y+20=0 ?

The solution to 2x^2+2y^2-16x+4y+20=0 is Circle with (a,b)=(4,-1),r=sqrt(7)

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2x^2+2y^2-16x+4y+20=0

2 x^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

(a, b) = (4, - 1), r = 7

Solution steps

2 x^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

Rewrite

2 x^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 1))^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (4, - 1), r = 7

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

x^{2} + y^{2} + 2 x - 8 y - 19 = 0

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

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

(x + 1)^{2} + y^{2} = 36

What is 2x^2+2y^2-16x+4y+20=0 ?
The solution to 2x^2+2y^2-16x+4y+20=0 is Circle with (a,b)=(4,-1),r=sqrt(7)