What is 2x^2+2y^2+6x-8y-3=0 ?

The solution to 2x^2+2y^2+6x-8y-3=0 is Circle with (a,b)=(-3/2 ,2),r=(sqrt(31))/2

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2x^2+2y^2+6x-8y-3=0

2 x^{2} + 2 y^{2} + 6 x - 8 y - 3 = 0

(a, b) = (- \frac{3}{2}, 2), r = \frac{31}{2}

Solution steps

2 x^{2} + 2 y^{2} + 6 x - 8 y - 3 = 0

Rewrite

2 x^{2} + 2 y^{2} + 6 x - 8 y - 3 = 0

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - 2)^{2} = (\frac{31}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 2), r = \frac{31}{2}

3 0^{2} = a^{2} + b^{2}

x^{2} + y^{2} + 6 x - 8 y + 5 = 0

(x + π)^{2} + (y - 2 π)^{2} = 8

18 y - 15 = - x^{2} - y^{2}

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

What is 2x^2+2y^2+6x-8y-3=0 ?
The solution to 2x^2+2y^2+6x-8y-3=0 is Circle with (a,b)=(-3/2 ,2),r=(sqrt(31))/2