What is the radius x^2+y^2-8x-13y-32=0 ?

The radius x^2+y^2-8x-13y-32=0 is r= 19/2

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radius x^2+y^2-8x-13y-32=0

radius

x^{2} + y^{2} - 8 x - 13 y - 32 = 0

r = \frac{19}{2}

Solution steps

x^{2} + y^{2} - 8 x - 13 y - 32 = 0

Rewrite

x^{2} + y^{2} - 8 x - 13 y - 32 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - \frac{13}{2})^{2} = (\frac{19}{2})^{2}

Therefore the circle properties are:

(a, b) = (4, \frac{13}{2}), r = \frac{19}{2}

And the radius is:

r = \frac{19}{2}

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

(x - 9)^{2} + y^{2} - 32 = 0

(x - 3)^{2} = 8 - (y + 3)^{2}

x^{2} + y^{2} + 10 y + 15 = 0

6 x^{2} + 6 y^{2} + 12 x - y = 0

What is the radius x^2+y^2-8x-13y-32=0 ?
The radius x^2+y^2-8x-13y-32=0 is r= 19/2