What is x^2+y^2-9x+4y-127=0 ?

The solution to x^2+y^2-9x+4y-127=0 is Circle with (a,b)=(9/2 ,-2),r=(11sqrt(5))/2

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x^2+y^2-9x+4y-127=0

x^{2} + y^{2} - 9 x + 4 y - 127 = 0

(a, b) = (\frac{9}{2}, - 2), r = \frac{11 5}{2}

Solution steps

x^{2} + y^{2} - 9 x + 4 y - 127 = 0

Rewrite

x^{2} + y^{2} - 9 x + 4 y - 127 = 0

in the form of the standard circle equation

(x - \frac{9}{2})^{2} + (y - (- 2))^{2} = (\frac{11 5}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{9}{2}, - 2), r = \frac{11 5}{2}

(x - 5)^{2} + (y - 4)^{2} = 36

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

x^{2} + y^{2} - 6 x - 18 y + 41 = 0

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

x^{2} + (y + \frac{3}{5})^{2} = 27

What is x^2+y^2-9x+4y-127=0 ?
The solution to x^2+y^2-9x+4y-127=0 is Circle with (a,b)=(9/2 ,-2),r=(11sqrt(5))/2