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

The solution to 2x^2+2y^2+3x+5y=5 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

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

(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

Rewrite

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

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^{2} - 12 x + y^{2} + 2 y - 12 = 0

x^{2} + y^{2} - 14 x + 8 y + 40 = 0

x^{2} + y^{2} + 10 x - 8 y + 37 = 0

(x - 5)^{2} + (y + 6)^{2} = 20

\frac{x ^{2}}{16} + \frac{y ^{2}}{16} = 1

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