What is the radius x^2+y^2+7x-2y-1=0 ?

The radius x^2+y^2+7x-2y-1=0 is r=(sqrt(57))/2

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radius x^2+y^2+7x-2y-1=0

radius

x^{2} + y^{2} + 7 x - 2 y - 1 = 0

r = \frac{57}{2}

Solution steps

x^{2} + y^{2} + 7 x - 2 y - 1 = 0

Rewrite

x^{2} + y^{2} + 7 x - 2 y - 1 = 0

in the form of the standard circle equation

(x - (- \frac{7}{2}))^{2} + (y - 1)^{2} = (\frac{57}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{2}, 1), r = \frac{57}{2}

And the radius is:

r = \frac{57}{2}

x^{2} + y^{2} + 18 x + 14 y + 105 = 0

(x + 3)^{2} + (y + 5)^{2} = 4

(x + 12)^{2} + (y - 13)^{2} - 25 = 0

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

(x + 1)^{2} + y^{2} = 246

What is the radius x^2+y^2+7x-2y-1=0 ?
The radius x^2+y^2+7x-2y-1=0 is r=(sqrt(57))/2