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

The radius 5x^2+5y^2+10x-5y+3=0 is r=(sqrt(13))/(2sqrt(5))

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

radius

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

r = \frac{13}{2 5}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{1}{2})^{2} = (\frac{13}{2 5})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{1}{2}), r = \frac{13}{2 5}

And the radius is:

r = \frac{13}{2 5}

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

x^{2} + y^{2} - 4 x + 2 y - 11 = 0

x^{2} + y^{2} - 6 x + 4 y = 7

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

x^{2} + y^{2} - 14 x + 6 y + 49 = 0

What is the radius 5x^2+5y^2+10x-5y+3=0 ?
The radius 5x^2+5y^2+10x-5y+3=0 is r=(sqrt(13))/(2sqrt(5))