What is x^2+4x+y^2-8y-5=0 ?

The solution to x^2+4x+y^2-8y-5=0 is Circle with (a,b)=(-2,4),r=5

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x^2+4x+y^2-8y-5=0

x^{2} + 4 x + y^{2} - 8 y - 5 = 0

(a, b) = (- 2, 4), r = 5

Solution steps

x^{2} + 4 x + y^{2} - 8 y - 5 = 0

Rewrite

x^{2} + 4 x + y^{2} - 8 y - 5 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 4), r = 5

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

\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

What is x^2+4x+y^2-8y-5=0 ?
The solution to x^2+4x+y^2-8y-5=0 is Circle with (a,b)=(-2,4),r=5