What is the radius 5x^2-10x+5y^2+3y-7=0 ?

The radius 5x^2-10x+5y^2+3y-7=0 is r=(sqrt(249))/(10)

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radius 5x^2-10x+5y^2+3y-7=0

radius

5 x^{2} - 10 x + 5 y^{2} + 3 y - 7 = 0

r = \frac{249}{10}

Solution steps

5 x^{2} - 10 x + 5 y^{2} + 3 y - 7 = 0

Rewrite

5 x^{2} - 10 x + 5 y^{2} + 3 y - 7 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- \frac{3}{10}))^{2} = (\frac{249}{10})^{2}

Therefore the circle properties are:

(a, b) = (1, - \frac{3}{10}), r = \frac{249}{10}

And the radius is:

r = \frac{249}{10}

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

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{9}{4}

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

x^{2} - 6 x + 14 y - 567 + y^{2} = 0

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

What is the radius 5x^2-10x+5y^2+3y-7=0 ?
The radius 5x^2-10x+5y^2+3y-7=0 is r=(sqrt(249))/(10)