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

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

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

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

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

Solution steps

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

Rewrite

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

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400

5 x^{2} + 7 y = 3

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

x^{2} + 14 x + y^{2} - 6 y - 23 = 0

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