What is 4x^2+6x+4y^2=7 ?

The solution to 4x^2+6x+4y^2=7 is Circle with (a,b)=(-3/4 ,0),r=(sqrt(37))/4

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4x^2+6x+4y^2=7

4 x^{2} + 6 x + 4 y^{2} = 7

(a, b) = (- \frac{3}{4}, 0), r = \frac{37}{4}

Solution steps

4 x^{2} + 6 x + 4 y^{2} = 7

Rewrite

4 x^{2} + 6 x + 4 y^{2} = 7

in the form of the standard circle equation

(x - (- \frac{3}{4}))^{2} + (y - 0)^{2} = (\frac{37}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, 0), r = \frac{37}{4}

\frac{x ^{2} + ( y - 6 ) ^{2}}{16} = 1

(x - 6)^{2} + (y - 3)^{2} = 4

(x - 5)^{2} + y^{2} = 81

y^{2} = 81 - x^{2}

(x - 0)^{2} + (y - 8)^{2} = 5^{2}

What is 4x^2+6x+4y^2=7 ?
The solution to 4x^2+6x+4y^2=7 is Circle with (a,b)=(-3/4 ,0),r=(sqrt(37))/4