What is 2x^2+2y^2-3x+4y-1=0 ?

The solution to 2x^2+2y^2-3x+4y-1=0 is Circle with (a,b)=(3/4 ,-1),r=(sqrt(33))/4

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2x^2+2y^2-3x+4y-1=0

2 x^{2} + 2 y^{2} - 3 x + 4 y - 1 = 0

(a, b) = (\frac{3}{4}, - 1), r = \frac{33}{4}

Solution steps

2 x^{2} + 2 y^{2} - 3 x + 4 y - 1 = 0

Rewrite

2 x^{2} + 2 y^{2} - 3 x + 4 y - 1 = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- 1))^{2} = (\frac{33}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - 1), r = \frac{33}{4}

x^{2} + y^{2} - 6 x - 8 y - 45 = 0

(x - 2)^{2} + (y - 1)^{2} = 1

2 x^{2} + 2 y^{2} - 50 = 0

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

y^{2} = - x^{2} + 25

What is 2x^2+2y^2-3x+4y-1=0 ?
The solution to 2x^2+2y^2-3x+4y-1=0 is Circle with (a,b)=(3/4 ,-1),r=(sqrt(33))/4