What is the radius 3x^2+5x+3y^2+6y-15=0 ?

The radius 3x^2+5x+3y^2+6y-15=0 is r=(sqrt(241))/6

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radius 3x^2+5x+3y^2+6y-15=0

radius

3 x^{2} + 5 x + 3 y^{2} + 6 y - 15 = 0

r = \frac{241}{6}

Solution steps

3 x^{2} + 5 x + 3 y^{2} + 6 y - 15 = 0

Rewrite

3 x^{2} + 5 x + 3 y^{2} + 6 y - 15 = 0

in the form of the standard circle equation

(x - (- \frac{5}{6}))^{2} + (y - (- 1))^{2} = (\frac{241}{6})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{6}, - 1), r = \frac{241}{6}

And the radius is:

r = \frac{241}{6}

x^{2} + y^{2} + 8 x - 14 y - 56 = 0

x^{2} + y^{2} + 3 x - 5 y + 1 = 0

x^{2} + y^{2} = 1 + 2 y

x^{2} + y^{2} - 8 \cdot + 7 = 0

(x + 2)^{2} + (y + 5)^{2} = 16

What is the radius 3x^2+5x+3y^2+6y-15=0 ?
The radius 3x^2+5x+3y^2+6y-15=0 is r=(sqrt(241))/6