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

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

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

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

(a, b) = (1, - \frac{3}{2}), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - \frac{3}{2}), r = 2

(x - 5)^{2} + (y + 2)^{2} = 15

(x + 1)^{2} + (y - 4)^{2} = 30

2 x^{2} + 2 y^{2} - 8 x - 16 y + 22 = 0

x^{2} + y^{2} + 4 y + 1 = 0

3 (x - 4)^{2} + 3 y^{2} = 27

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