What is x^2+y^2-9x+4y-117=0 ?

The solution to x^2+y^2-9x+4y-117=0 is Circle with (a,b)=(9/2 ,-2),r=(sqrt(565))/2

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x^2+y^2-9x+4y-117=0

x^{2} + y^{2} - 9 x + 4 y - 117 = 0

(a, b) = (\frac{9}{2}, - 2), r = \frac{565}{2}

Solution steps

x^{2} + y^{2} - 9 x + 4 y - 117 = 0

Rewrite

x^{2} + y^{2} - 9 x + 4 y - 117 = 0

in the form of the standard circle equation

(x - \frac{9}{2})^{2} + (y - (- 2))^{2} = (\frac{565}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{9}{2}, - 2), r = \frac{565}{2}

y^{2} + x^{2} - 16 x + 16 y + 92 = 0

x^{2} + y^{2} + 12 x - 16 y = 0

a^{2} + b^{2} = 6

(x - 1)^{2} + (y - 5)^{2} = 81

x^{2} + (y - 22.5)^{2} = 22. 5^{2}

What is x^2+y^2-9x+4y-117=0 ?
The solution to x^2+y^2-9x+4y-117=0 is Circle with (a,b)=(9/2 ,-2),r=(sqrt(565))/2