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