What is 4x^2+12x+9+4y^2>7 ?

The solution to 4x^2+12x+9+4y^2>7 is Circle with (a,b)=(-3/2 ,0),r=(sqrt(7))/2

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4x^2+12x+9+4y^2>7

4 x^{2} + 12 x + 9 + 4 y^{2} > 7

(a, b) = (- \frac{3}{2}, 0), r = \frac{7}{2}

Solution steps

4 x^{2} + 12 x + 9 + 4 y^{2} > 7

Rewrite

4 x^{2} + 12 x + 9 + 4 y^{2} > 7

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - 0)^{2} = (\frac{7}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 0), r = \frac{7}{2}

x^{2} + y^{2} - 2 x - 8 y + 1 = 0

(x - 5)^{2} + (Y (x) + 2)^{2} = 16

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

2 x^{2} + 12 x + 2 y^{2} - 2 y - 3 = 0

x^{2} + y^{2} + 6 x - 13 y + 13 = 0

What is 4x^2+12x+9+4y^2>7 ?
The solution to 4x^2+12x+9+4y^2>7 is Circle with (a,b)=(-3/2 ,0),r=(sqrt(7))/2