What is 2x^2+2y^2+8x-6y+4=0 ?

The solution to 2x^2+2y^2+8x-6y+4=0 is Circle with (a,b)=(-2, 3/2),r=(sqrt(17))/2

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2x^2+2y^2+8x-6y+4=0

2 x^{2} + 2 y^{2} + 8 x - 6 y + 4 = 0

(a, b) = (- 2, \frac{3}{2}), r = \frac{17}{2}

Solution steps

2 x^{2} + 2 y^{2} + 8 x - 6 y + 4 = 0

Rewrite

2 x^{2} + 2 y^{2} + 8 x - 6 y + 4 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - \frac{3}{2})^{2} = (\frac{17}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, \frac{3}{2}), r = \frac{17}{2}

(x - 5)^{2} + (y + 2)^{2} = 9

x^{2} + y^{2} = 85

(x - 1)^{2} + (y - 2)^{2} = 81

a^{2} + b^{2} = 64

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

What is 2x^2+2y^2+8x-6y+4=0 ?
The solution to 2x^2+2y^2+8x-6y+4=0 is Circle with (a,b)=(-2, 3/2),r=(sqrt(17))/2