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

The radius x^2+y^2-9x+4y-127=0 is r=(11sqrt(5))/2

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

radius

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

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}

And the radius is:

r = \frac{11 5}{2}

x^{2} + y^{2} + 14 x - 18 y + 49 = 0

4 x^{2} + 4 y^{2} - 8 x + 4 y - 11 = 0

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

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

What is the radius x^2+y^2-9x+4y-127=0 ?
The radius x^2+y^2-9x+4y-127=0 is r=(11sqrt(5))/2