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

The solution to 2x^2+2y^2+x-6=0 is Circle with (a,b)=(-1/4 ,0),r= 7/4

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

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

(a, b) = (- \frac{1}{4}, 0), r = \frac{7}{4}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{4}))^{2} + (y - 0)^{2} = (\frac{7}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{4}, 0), r = \frac{7}{4}

x^{2} + y^{2} - 2 x + 2 y - 23 = 0

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

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

x^{2} + y^{2} - x + 6 y + 1 = 0

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

What is 2x^2+2y^2+x-6=0 ?
The solution to 2x^2+2y^2+x-6=0 is Circle with (a,b)=(-1/4 ,0),r= 7/4