What is x^2+y^2+520x+160y+202=0 ?

The solution to x^2+y^2+520x+160y+202=0 is Circle with (a,b)=(-260,-80),r=sqrt(73798)

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x^2+y^2+520x+160y+202=0

x^{2} + y^{2} + 520 x + 160 y + 202 = 0

(a, b) = (- 260, - 80), r = 73798

Solution steps

x^{2} + y^{2} + 520 x + 160 y + 202 = 0

Rewrite

x^{2} + y^{2} + 520 x + 160 y + 202 = 0

in the form of the standard circle equation

(x - (- 260))^{2} + (y - (- 80))^{2} = (73798)^{2}

Therefore the circle properties are:

(a, b) = (- 260, - 80), r = 73798

2 x^{2} + 2 y^{2} + 20 x + 8 y + 40 = 0

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

x, x^{2} - 6 x + Y (x)^{2} + 4 Y = 36

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

1 = 16 - x^{2} - y^{2}

What is x^2+y^2+520x+160y+202=0 ?
The solution to x^2+y^2+520x+160y+202=0 is Circle with (a,b)=(-260,-80),r=sqrt(73798)