What is 2x^2+2y^2+16x+20y+18=0 ?

The solution to 2x^2+2y^2+16x+20y+18=0 is Circle with (a,b)=(-4,-5),r=4sqrt(2)

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2x^2+2y^2+16x+20y+18=0

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

(a, b) = (- 4, - 5), r = 42

Solution steps

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

Rewrite

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 5))^{2} = (42)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 5), r = 42

\frac{x ^{2}}{2} + \frac{y ^{2}}{2} = 2

x^{2} - 4 x + y^{2} - 9 x + 9 = 0

0 = 6 - 4 (x^{2} + y^{2})

x^{2} + y^{2} + 16 x + 14 y + 88 = 0

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

What is 2x^2+2y^2+16x+20y+18=0 ?
The solution to 2x^2+2y^2+16x+20y+18=0 is Circle with (a,b)=(-4,-5),r=4sqrt(2)