What is x^2+y^2-9x+4y-13=0 ?

The solution to x^2+y^2-9x+4y-13=0 is Circle with (a,b)=(9/2 ,-2),r=(sqrt(149))/2

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x^2+y^2-9x+4y-13=0

x^{2} + y^{2} - 9 x + 4 y - 13 = 0

(a, b) = (\frac{9}{2}, - 2), r = \frac{149}{2}

Solution steps

x^{2} + y^{2} - 9 x + 4 y - 13 = 0

Rewrite

x^{2} + y^{2} - 9 x + 4 y - 13 = 0

in the form of the standard circle equation

(x - \frac{9}{2})^{2} + (y - (- 2))^{2} = (\frac{149}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{9}{2}, - 2), r = \frac{149}{2}

(x - 15)^{2} + (y - 2)^{2} = 9

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

(x - 4)^{2} + (y - 3)^{2} = 2 [(x + 2)^{2} + (y - 1)^{2}]

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

(x - 37.5)^{2} + (y + 2.5)^{2} = 1^{2}

What is x^2+y^2-9x+4y-13=0 ?
The solution to x^2+y^2-9x+4y-13=0 is Circle with (a,b)=(9/2 ,-2),r=(sqrt(149))/2