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

The solution to x^2+y^2-2x+4y-4=0 is Circle with (a,b)=(1,-2),r=3

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

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

(a, b) = (1, - 2), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 2))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (1, - 2), r = 3

x = - \frac{1}{12} y^{2}

4 x^{2} + 9 y^{2} - 16 x + 18 y = 11

\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1

x^{2}

y^{2} = - 12 x

What is x^2+y^2-2x+4y-4=0 ?
The solution to x^2+y^2-2x+4y-4=0 is Circle with (a,b)=(1,-2),r=3