What is 2x^2+2y^2+3x+5y-5=0 ?

The solution to 2x^2+2y^2+3x+5y-5=0 is Circle with (a,b)=(-3/4 ,-5/4),r=(sqrt(37))/(2sqrt(2))

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2x^2+2y^2+3x+5y-5=0

2 x^{2} + 2 y^{2} + 3 x + 5 y - 5 = 0

(a, b) = (- \frac{3}{4}, - \frac{5}{4}), r = \frac{37}{2 2}

Solution steps

2 x^{2} + 2 y^{2} + 3 x + 5 y - 5 = 0

Rewrite

2 x^{2} + 2 y^{2} + 3 x + 5 y - 5 = 0

in the form of the standard circle equation

(x - (- \frac{3}{4}))^{2} + (y - (- \frac{5}{4}))^{2} = (\frac{37}{2 2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, - \frac{5}{4}), r = \frac{37}{2 2}

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

x^{2} + y^{2} - 6 x - 8 y - 17 = 0

x^{2} + y^{2} + 2 x + 4 y - 20 = 0

x^{2} + y^{2} - 6 y - 1 = 0

(x - 4)^{2} + (y + 2)^{2} = 16

What is 2x^2+2y^2+3x+5y-5=0 ?
The solution to 2x^2+2y^2+3x+5y-5=0 is Circle with (a,b)=(-3/4 ,-5/4),r=(sqrt(37))/(2sqrt(2))