What is x^2-3x+y^2+1=0 ?

The solution to x^2-3x+y^2+1=0 is Circle with (a,b)=(3/2 ,0),r=(sqrt(5))/2

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x^2-3x+y^2+1=0

x^{2} - 3 x + y^{2} + 1 = 0

(a, b) = (\frac{3}{2}, 0), r = \frac{5}{2}

Solution steps

x^{2} - 3 x + y^{2} + 1 = 0

Rewrite

x^{2} - 3 x + y^{2} + 1 = 0

in the form of the standard circle equation

(x - \frac{3}{2})^{2} + (y - 0)^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{2}, 0), r = \frac{5}{2}

x^{2} + y^{2} + 4 x - 10 y - 7 = 0

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

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

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

16 x^{2} + 80 x + 16 y^{2} - 112 y + 247 = 0

What is x^2-3x+y^2+1=0 ?
The solution to x^2-3x+y^2+1=0 is Circle with (a,b)=(3/2 ,0),r=(sqrt(5))/2