What is 2x^2+2y^2-3x+y-1=0 ?

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

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2x^2+2y^2-3x+y-1=0

2 x^{2} + 2 y^{2} - 3 x + y - 1 = 0

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

Solution steps

2 x^{2} + 2 y^{2} - 3 x + y - 1 = 0

Rewrite

2 x^{2} + 2 y^{2} - 3 x + y - 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

3 x^{2} + 3 y^{2} = 3

(x - 5)^{2} + (y - 5)^{2} = 100

x^{2} + y^{2} + \frac{20}{17} x - \frac{72}{17} y - \frac{912}{17} = 0

x^{2} + y^{2} + 4 x - 4 y - 32 = 0

72 = x^{2} + 6 x + 9 + y^{2} - 4 y + 4

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