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

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

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

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

(a, b) = (- 4, - 2), r = 26

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 2))^{2} = (26)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 2), r = 26

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