What is the radius 2x^2+2y^2-3x+4y-35=0 ?

The radius 2x^2+2y^2-3x+4y-35=0 is r=(sqrt(305))/4

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radius 2x^2+2y^2-3x+4y-35=0

radius

2 x^{2} + 2 y^{2} - 3 x + 4 y - 35 = 0

r = \frac{305}{4}

Solution steps

2 x^{2} + 2 y^{2} - 3 x + 4 y - 35 = 0

Rewrite

2 x^{2} + 2 y^{2} - 3 x + 4 y - 35 = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- 1))^{2} = (\frac{305}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - 1), r = \frac{305}{4}

And the radius is:

r = \frac{305}{4}

(x - 5)^{2} + (y - 3)^{2} = 64

x^{2} + (y + 2)^{2} = 36

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

(x - 3)^{2} + (y + 4)^{2} = 25

x^{2} + y^{2} + 16 = 9 x

What is the radius 2x^2+2y^2-3x+4y-35=0 ?
The radius 2x^2+2y^2-3x+4y-35=0 is r=(sqrt(305))/4