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

The solution to 2x^2+2y^2+3*+5y-5=0 is Circle with (a,b)=(0,-5/4),r=(sqrt(41))/4

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

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

(a, b) = (0, - \frac{5}{4}), r = \frac{41}{4}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{5}{4}))^{2} = (\frac{41}{4})^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{5}{4}), r = \frac{41}{4}

x^{2} + y^{2} - 8 x + 2 y = - 1

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

2 x^{2} + 2 y^{2} + 4 x - 8 y - 40 = 0

x^{2} + (y + 7)^{2} = 36

x^{2} + y^{2} - 8 x - 20 y + 67 = 0

What is 2x^2+2y^2+3*+5y-5=0 ?
The solution to 2x^2+2y^2+3*+5y-5=0 is Circle with (a,b)=(0,-5/4),r=(sqrt(41))/4