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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 2 x - 15 + 14 y = 0

(x^{2} + 3 x) + (y^{2} - 4 y) = 6

2 x^{2} + 2 y^{2} - 2 x + 4 y = 45

2 x^{(2)} + 2 y^{(2)} - 4 x + 12 y - 12 = 0

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

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