What is x^2+y^2+10x+4y-56=0 ?

The solution to x^2+y^2+10x+4y-56=0 is Circle with (a,b)=(-5,-2),r=sqrt(85)

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x^2+y^2+10x+4y-56=0

x^{2} + y^{2} + 10 x + 4 y - 56 = 0

(a, b) = (- 5, - 2), r = 85

Solution steps

x^{2} + y^{2} + 10 x + 4 y - 56 = 0

Rewrite

x^{2} + y^{2} + 10 x + 4 y - 56 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 2))^{2} = (85)^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 2), r = 85

x^{2} + y^{2} - 5 x = 44 - 2 y

(x^{2} + 4 x) + (y^{2} + 4 y) = 9

x^{2} + y^{2} + 12 x + 2 y + \frac{1}{16} = 0

(x - 3)^{2} + (y + 2)^{2} = 17

x^{2} + y^{2} - 12 x = - 4 y + 5

What is x^2+y^2+10x+4y-56=0 ?
The solution to x^2+y^2+10x+4y-56=0 is Circle with (a,b)=(-5,-2),r=sqrt(85)