What is 2x^2+2y^2-5x+4y-1=0 ?

The solution to 2x^2+2y^2-5x+4y-1=0 is Circle with (a,b)=(5/4 ,-1),r= 7/4

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2x^2+2y^2-5x+4y-1=0

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

(a, b) = (\frac{5}{4}, - 1), r = \frac{7}{4}

Solution steps

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

Rewrite

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

in the form of the standard circle equation

(x - \frac{5}{4})^{2} + (y - (- 1))^{2} = (\frac{7}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{4}, - 1), r = \frac{7}{4}

x^{2} + y^{2} - 3 y = 12

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

x^{2} + y^{2} + 2 x - 6 y - 15 = 0

x^{2} + y^{2} + 4 y - 5 = 0

x^{2} + (y - 5)^{2} = 36

What is 2x^2+2y^2-5x+4y-1=0 ?
The solution to 2x^2+2y^2-5x+4y-1=0 is Circle with (a,b)=(5/4 ,-1),r= 7/4