What is the radius x^2+y^2+7x-4y-2=0 ?

The radius x^2+y^2+7x-4y-2=0 is r=(sqrt(73))/2

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radius x^2+y^2+7x-4y-2=0

radius

x^{2} + y^{2} + 7 x - 4 y - 2 = 0

r = \frac{73}{2}

Solution steps

x^{2} + y^{2} + 7 x - 4 y - 2 = 0

Rewrite

x^{2} + y^{2} + 7 x - 4 y - 2 = 0

in the form of the standard circle equation

(x - (- \frac{7}{2}))^{2} + (y - 2)^{2} = (\frac{73}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{7}{2}, 2), r = \frac{73}{2}

And the radius is:

r = \frac{73}{2}

y^{2} + x^{2} = 25

x^{2} + y^{2} - 14 x - 12 y + 59 = 0

(x + 5)^{2} + (y - 2)^{2} = 25

x^{2} + (y - 4)^{2} = 1

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

What is the radius x^2+y^2+7x-4y-2=0 ?
The radius x^2+y^2+7x-4y-2=0 is r=(sqrt(73))/2