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

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

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

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

(a, b) = (1, 0), r = \frac{3}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 1)^{2} + (y - 0)^{2} = (\frac{3}{2})^{2}

Therefore the circle properties are:

(a, b) = (1, 0), r = \frac{3}{2}

\frac{x ^{2}}{1} + \frac{y ^{2}}{1} = 1

(x + 2)^{2} + (y + 2)^{2} = 4

(x + 6)^{2} + (y - 10)^{2} = 12

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

x^{2} + y^{2} - 9 x + 4 y - 138 = 0

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