What is x^2+y^2+1x+5y-56=0 ?

The solution to x^2+y^2+1x+5y-56=0 is Circle with (a,b)=(-1/2 ,-5/2),r=(5sqrt(5))/(sqrt(2))

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x^2+y^2+1x+5y-56=0

x^{2} + y^{2} + 1 x + 5 y - 56 = 0

(a, b) = (- \frac{1}{2}, - \frac{5}{2}), r = \frac{5 5}{2}

Solution steps

x^{2} + y^{2} + 1 \cdot x + 5 y - 56 = 0

Rewrite

x^{2} + y^{2} + 1 x + 5 y - 56 = 0

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - (- \frac{5}{2}))^{2} = (\frac{5 5}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, - \frac{5}{2}), r = \frac{5 5}{2}

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} - \frac{1}{8} = 0

x^{2} + y^{2} - 16 x - 2 y + 65 = 64

x^{2} - 8 x - 16 + y^{2} + 2 y + 1 = 25

(x - 2)^{2} + (y + 3)^{2} = 100

2 x^{2} + 2 y^{2} + 3 x + 5 y = 5

What is x^2+y^2+1x+5y-56=0 ?
The solution to x^2+y^2+1x+5y-56=0 is Circle with (a,b)=(-1/2 ,-5/2),r=(5sqrt(5))/(sqrt(2))