What is x^2-4x+y^2+3y=0 ?

The solution to x^2-4x+y^2+3y=0 is Circle with (a,b)=(2,-3/2),r= 5/2

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x^2-4x+y^2+3y=0

x^{2} - 4 x + y^{2} + 3 y = 0

(a, b) = (2, - \frac{3}{2}), r = \frac{5}{2}

Solution steps

x^{2} - 4 x + y^{2} + 3 y = 0

Rewrite

x^{2} - 4 x + y^{2} + 3 y = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (2, - \frac{3}{2}), r = \frac{5}{2}

2 x^{2} + 2 y^{2} + 3 x - 2 y - 43 = 0

x^{2} + y^{2} - 2 x + 4 y + 8 = 7

x^{2} + y^{2} + 4 x - 12 y - 3 = 0

2 x^{2} - 8 x + 8 + 2 y^{2} + 4 x + 2 = 25

100 = x^{2} + y^{2} + 4 x + 60

What is x^2-4x+y^2+3y=0 ?
The solution to x^2-4x+y^2+3y=0 is Circle with (a,b)=(2,-3/2),r= 5/2