What is 2x^2+3x+2y^2+7y=19 ?

The solution to 2x^2+3x+2y^2+7y=19 is Circle with (a,b)=(-3/4 ,-7/4),r=(sqrt(105))/(2sqrt(2))

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2x^2+3x+2y^2+7y=19

2 x^{2} + 3 x + 2 y^{2} + 7 y = 19

(a, b) = (- \frac{3}{4}, - \frac{7}{4}), r = \frac{105}{2 2}

Solution steps

2 x^{2} + 3 x + 2 y^{2} + 7 y = 19

Rewrite

2 x^{2} + 3 x + 2 y^{2} + 7 y = 19

in the form of the standard circle equation

(x - (- \frac{3}{4}))^{2} + (y - (- \frac{7}{4}))^{2} = (\frac{105}{2 2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, - \frac{7}{4}), r = \frac{105}{2 2}

(x + 4)^{2} + (y - 1)^{2} = 148

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

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

x^{2} + (y + 6)^{2} = 50

(x + 3)^{2} + (y - 1)^{2} = 32

What is 2x^2+3x+2y^2+7y=19 ?
The solution to 2x^2+3x+2y^2+7y=19 is Circle with (a,b)=(-3/4 ,-7/4),r=(sqrt(105))/(2sqrt(2))