Answer
$\left(\frac{3-3\sqrt{7}}{2},\frac{-3 -3\sqrt{7}}{2}\right),\left(\frac{3+3\sqrt{7}}{2},\frac{-3+3\sqrt{7}}{2}\right)$
Work Step by Step
Substituting the value of $y$ from the second equation into the first equation, we obtain
$$
\begin{aligned}
x^2+(x-3)^2 & =36 \\
x^2+x^2-6 x+9 & =36 \\
2 x^2-6 x-27 & =0
\end{aligned}
$$
$$
\begin{aligned}
x & =\frac{-(-6) \pm \sqrt{(-6)^2-4(2)(-27)}}{(2)(2)} \\
& =\frac{6 \pm \sqrt{252}}{4} \\
& =\frac{6 \pm \sqrt{4 \cdot 9\cdot 7}}{4} \\
& =\frac{6 \pm 6 \sqrt{7}}{4} \\
& =\frac{3 \pm 3\sqrt{7}}{2} .
\end{aligned}
$$
Now we substitute the values of $x$ into the second equation:
$$
\begin{aligned}
x_1&=\frac{3-3\sqrt{7}}{2}\\
y_1 & =\frac{3-3\sqrt{7}}{2}-3 \\
& =\frac{-3 -3\sqrt{7}}{2}\\
x_2&=\frac{3+3\sqrt{7}}{2}\\
y_2 & =\frac{3+3\sqrt{7}}{2}-3 \\
& =\frac{-3 +3\sqrt{7}}{2}
\end{aligned}
$$
The solutions are:
$$\left(\frac{3-3\sqrt{7}}{2},\frac{-3 -3\sqrt{7}}{2}\right),\left(\frac{3+3\sqrt{7}}{2},\frac{-3+3\sqrt{7}}{2}\right)$$
