What is 2x^2+2y^2-3x+2y-10=0 ?

The solution to 2x^2+2y^2-3x+2y-10=0 is Circle with (a,b)=(3/4 ,-1/2),r=(sqrt(93))/4

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2x^2+2y^2-3x+2y-10=0

2 x^{2} + 2 y^{2} - 3 x + 2 y - 10 = 0

(a, b) = (\frac{3}{4}, - \frac{1}{2}), r = \frac{93}{4}

Solution steps

2 x^{2} + 2 y^{2} - 3 x + 2 y - 10 = 0

Rewrite

2 x^{2} + 2 y^{2} - 3 x + 2 y - 10 = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- \frac{1}{2}))^{2} = (\frac{93}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - \frac{1}{2}), r = \frac{93}{4}

(x - 3)^{2} + (y + 4)^{2} = 36

8 x + 32 y + y^{2} = - 263 - x^{2}

2 x^{2} + 2 y^{2} - 8 x + 28 y + 86 = 0

(x + 12)^{2} + (Y (x) - 4)^{2} - 30 = 0

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

What is 2x^2+2y^2-3x+2y-10=0 ?
The solution to 2x^2+2y^2-3x+2y-10=0 is Circle with (a,b)=(3/4 ,-1/2),r=(sqrt(93))/4