What is the radius 2x^2+2y^2+16x-6y-18=0 ?

The radius 2x^2+2y^2+16x-6y-18=0 is r=(sqrt(109))/2

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radius 2x^2+2y^2+16x-6y-18=0

radius

2 x^{2} + 2 y^{2} + 16 x - 6 y - 18 = 0

r = \frac{109}{2}

Solution steps

2 x^{2} + 2 y^{2} + 16 x - 6 y - 18 = 0

Rewrite

2 x^{2} + 2 y^{2} + 16 x - 6 y - 18 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - \frac{3}{2})^{2} = (\frac{109}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 4, \frac{3}{2}), r = \frac{109}{2}

And the radius is:

r = \frac{109}{2}

(x - (- 5))^{2} + (y - 7)^{2} = 3^{2}

(x + 6)^{2} + (y - 1)^{2} = 16

x^{2} + y^{2} - 7 x = 15 - 2 y

(x - 3)^{2} + (y + 5)^{2} = 25

x^{2} + (y + 4)^{2} = 121

What is the radius 2x^2+2y^2+16x-6y-18=0 ?
The radius 2x^2+2y^2+16x-6y-18=0 is r=(sqrt(109))/2