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}

4 x^{2} + 4 y^{2} - 24 x + 8 y + 39 = 0

(x - 4)^{2} + (y - 5)^{2} = 64

(x)^{2} + (y)^{2} - 6 x + 10 y + 34 = 72

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

2 x^{2} - 4 x + 2 y^{2} + 18 y = 7

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