What is 2x^2+12x+18-y^2=3(2-y^2)+4y ?

The solution to 2x^2+12x+18-y^2=3(2-y^2)+4y is Circle with (a,b)=(-3,1),r=2

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2x^2+12x+18-y^2=3(2-y^2)+4y

2 x^{2} + 12 x + 18 - y^{2} = 3 (2 - y^{2}) + 4 y

(a, b) = (- 3, 1), r = 2

Solution steps

2 x^{2} + 12 x + 18 - y^{2} = 3 (2 - y^{2}) + 4 y

Rewrite

2 x^{2} + 12 x + 18 - y^{2} = 3 (2 - y^{2}) + 4 y

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 3, 1), r = 2

x^{2} + y^{2} - x \geq 0

x^{2} + y^{2} = 196

2 x^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

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

x^{2} + y^{2} + 2 x - 8 y - 19 = 0

What is 2x^2+12x+18-y^2=3(2-y^2)+4y ?
The solution to 2x^2+12x+18-y^2=3(2-y^2)+4y is Circle with (a,b)=(-3,1),r=2