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

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

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

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

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

Solution steps

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

Rewrite

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

x^{2} + y^{2} - 6 x + 4 y - 11 = 0

74 = (24 - x)^{2} + (22 - y)^{2}

(x + 1)^{2} + (y - 1)^{2} = 9 ((x - 2)^{2} + (y + 1)^{2})

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

(x - 4)^{2} + y^{2} \leq 4

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