1. **State the problem:** Solve the system of equations: $$x^2 + y^2 = 2$$ $$x - y = 1$$ 2. **Use substitution:** From the second equation, express $x$ in terms of $y$: $$x = y + 1$$ 3. **Substitute into the first equation:** Replace $x$ with $y + 1$ in the circle equation: $$ (y + 1)^2 + y^2 = 2 $$ 4. **Expand and simplify:** $$ y^2 + 2y + 1 + y^2 = 2 $$ $$ 2y^2 + 2y + 1 = 2 $$ 5. **Bring all terms to one side:** $$ 2y^2 + 2y + 1 - 2 = 0 $$ $$ 2y^2 + 2y - 1 = 0 $$ 6. **Divide entire equation by 2 to simplify:** $$ \cancel{2}y^2 + \cancel{2}y - \cancel{2} \frac{1}{2} = 0 $$ $$ y^2 + y - \frac{1}{2} = 0 $$ 7. **Solve quadratic equation using the quadratic formula:** $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=1$, $b=1$, $c=-\frac{1}{2}$. Calculate discriminant: $$ \Delta = 1^2 - 4 \times 1 \times \left(-\frac{1}{2}\right) = 1 + 2 = 3 $$ So, $$ y = \frac{-1 \pm \sqrt{3}}{2} $$ 8. **Find corresponding $x$ values:** Recall $x = y + 1$: $$ x = \frac{-1 \pm \sqrt{3}}{2} + 1 = \frac{-1 \pm \sqrt{3} + 2}{2} = \frac{1 \pm \sqrt{3}}{2} $$ 9. **Final solutions:** $$ \left( \frac{1 + \sqrt{3}}{2}, \frac{-1 + \sqrt{3}}{2} \right) \quad \text{and} \quad \left( \frac{1 - \sqrt{3}}{2}, \frac{-1 - \sqrt{3}}{2} \right) $$ These are the points where the line intersects the circle.