Question 314735
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x^2\ +\ 9x^2\ -\ 6x\ +\ 12y\ -\ 31\ =\ 0]


is most emphatically not a circle.  Can't be a circle without a y-squared term.  However, I suspect you made a little typo and really meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x^2\ +\ 9y^2\ -\ 6x\ +\ 12y\ -\ 31\ =\ 0]


Complete the square on x:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\left(x^2\ -\ \frac{2}{3}x\ +\ \frac{1}{9}\right)\ +\ 9y^2\ +\ 12y\ -\ 31\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\left(x\ -\ \frac{1}{3}\right)^2\ +\ 9y^2\ +\ 12y\ -\ 31\ =\ 1]


Complete the square on y:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\left(x\ -\ \frac{1}{3}\right)^2+\ 9\left(y^2\ +\ \frac{4}{3}y\ +\ \frac{4}{9}\right)\ -\ 31\ =\ 1\ +\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\left(x\ -\ \frac{1}{3}\right)^2+\ 9\left(y\  +\ \frac{2}{3}\right)^2\ -\ 31\ =\ 1\ +\ 4]


And put the rest of the constant in the RHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\left(x\ -\ \frac{1}{3}\right)^2+\ 9\left(y\  +\ \frac{2}{3}\right)^2\ =\ 36]


Multiply both sides by *[tex \Large \frac{1}{9}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \frac{1}{3}\right)^2+\ \left(y\  +\ \frac{2}{3}\right)^2\ =\ 4]


Circle, center at *[tex \Large \left(\frac{1}{3},-\frac{2}{3}\right)] and radius 2.


Substitute zero for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(0\ -\ \frac{1}{3}\right)^2+\ \left(y\  +\ \frac{2}{3}\right)^2\ =\ 4]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\  +\ \frac{2}{3}\right)^2\ =\ \frac{35}{9}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\  +\ \frac{2}{3}\right)\ =\ \pm\frac{\sqrt{35}}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{2}{3}\ \pm\ \frac{\sqrt{35}}{3}]


Hence, the *[tex \Large x] intercepts are:  *[tex \Large \left(0,\frac{-2\ +\ \sqrt{35}}{3}\right)] and *[tex \Large \left(0,\frac{-2\ -\ \sqrt{35}}{3}\right)]


Just substitute 0 for y in the circle equation and do the arithmetic as above to find the  *[tex \Large y]-intercepts.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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