1. **State the problem:** We are given the system of equations: $$ (x-1)^2 + y^2 = 2 $$ and $$ y = x $$ We need to find the points $(x,y)$ that satisfy both equations. 2. **Substitute the second equation into the first:** Since $y = x$, replace $y$ in the first equation: $$ (x-1)^2 + x^2 = 2 $$ 3. **Expand and simplify:** $$ (x-1)^2 + x^2 = (x^2 - 2x + 1) + x^2 = 2x^2 - 2x + 1 $$ Set equal to 2: $$ 2x^2 - 2x + 1 = 2 $$ 4. **Bring all terms to one side:** $$ 2x^2 - 2x + 1 - 2 = 0 $$ $$ 2x^2 - 2x - 1 = 0 $$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=2$, $b=-2$, $c=-1$. Calculate the discriminant: $$ \Delta = (-2)^2 - 4 \times 2 \times (-1) = 4 + 8 = 12 $$ So, $$ x = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2} $$ 6. **Find corresponding $y$ values:** Since $y = x$, the points are: $$ \left( \frac{1 + \sqrt{3}}{2}, \frac{1 + \sqrt{3}}{2} \right) \quad \text{and} \quad \left( \frac{1 - \sqrt{3}}{2}, \frac{1 - \sqrt{3}}{2} \right) $$ **Final answer:** The solutions to the system are $$ \boxed{\left( \frac{1 + \sqrt{3}}{2}, \frac{1 + \sqrt{3}}{2} \right), \left( \frac{1 - \sqrt{3}}{2}, \frac{1 - \sqrt{3}}{2} \right)} $$