What is x^2-4x+y^2-9x+9=0 ?

The solution to x^2-4x+y^2-9x+9=0 is Circle with (a,b)=(13/2 ,0),r=(sqrt(133))/2

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x^2-4x+y^2-9x+9=0

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

(a, b) = (\frac{13}{2}, 0), r = \frac{133}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - \frac{13}{2})^{2} + (y - 0)^{2} = (\frac{133}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{13}{2}, 0), r = \frac{133}{2}

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

x^{2} + y^{2} + 16 x + 14 y + 88 = 0

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

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

4 x^{2} + 4 y^{2} - 16 x - 32 y = 0

What is x^2-4x+y^2-9x+9=0 ?
The solution to x^2-4x+y^2-9x+9=0 is Circle with (a,b)=(13/2 ,0),r=(sqrt(133))/2