What is x^2-2x+y^2+4y-20=0 ?

The solution to x^2-2x+y^2+4y-20=0 is Circle with (a,b)=(1,-2),r=5

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x^2-2x+y^2+4y-20=0

x^{2} - 2 x + y^{2} + 4 y - 20 = 0

(a, b) = (1, - 2), r = 5

Solution steps

x^{2} - 2 x + y^{2} + 4 y - 20 = 0

Rewrite

x^{2} - 2 x + y^{2} + 4 y - 20 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 2), r = 5

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

4 x^{2} + 16 x + 15 + 4 y^{2} + 6 y = 0

x^{2} + y^{2} - 2 y - 8 x - 19 = 0

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

(x^{2} - 4 x) + (y^{2} - 6 y) = - 9

What is x^2-2x+y^2+4y-20=0 ?
The solution to x^2-2x+y^2+4y-20=0 is Circle with (a,b)=(1,-2),r=5