What is the radius x^2+y^2+3x-4y-15=0 ?

The radius x^2+y^2+3x-4y-15=0 is r=(sqrt(85))/2

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radius x^2+y^2+3x-4y-15=0

radius

x^{2} + y^{2} + 3 x - 4 y - 15 = 0

r = \frac{85}{2}

Solution steps

x^{2} + y^{2} + 3 x - 4 y - 15 = 0

Rewrite

x^{2} + y^{2} + 3 x - 4 y - 15 = 0

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - 2)^{2} = (\frac{85}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 2), r = \frac{85}{2}

And the radius is:

r = \frac{85}{2}

(x + 6)^{2} + y^{2} = 1

(x + 2)^{2} + (y - 6)^{2} = 169

x^{2} + y^{2} - 2 x - 8 y - 19 = 0 (1.5) 6

x^{2} + y^{2} - 10 x + 12 y = 195

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

What is the radius x^2+y^2+3x-4y-15=0 ?
The radius x^2+y^2+3x-4y-15=0 is r=(sqrt(85))/2