1. **State the problem:** Find the coordinates of the points of intersection of the line $x - y = 1$ and the circle $x^2 + y^2 = 13$. 2. **Rewrite the line equation:** From $x - y = 1$, express $x$ in terms of $y$: $$x = y + 1$$ 3. **Substitute into the circle equation:** Replace $x$ in $x^2 + y^2 = 13$ with $y + 1$: $$ (y + 1)^2 + y^2 = 13 $$ 4. **Expand and simplify:** $$ y^2 + 2y + 1 + y^2 = 13 $$ $$ 2y^2 + 2y + 1 = 13 $$ $$ 2y^2 + 2y + 1 - 13 = 0 $$ $$ 2y^2 + 2y - 12 = 0 $$ 5. **Divide entire equation by 2 to simplify:** $$ \cancel{2}y^2 + \cancel{2}y - \cancel{12} = 0 \Rightarrow y^2 + y - 6 = 0 $$ 6. **Use the quadratic formula to solve for $y$:** The quadratic formula is: $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=1$, $b=1$, and $c=-6$. 7. **Calculate the discriminant:** $$ b^2 - 4ac = 1^2 - 4(1)(-6) = 1 + 24 = 25 $$ 8. **Find the roots:** $$ y = \frac{-1 \pm \sqrt{25}}{2(1)} = \frac{-1 \pm 5}{2} $$ So, - For $+$ sign: $$ y = \frac{-1 + 5}{2} = \frac{4}{2} = 2 $$ - For $-$ sign: $$ y = \frac{-1 - 5}{2} = \frac{-6}{2} = -3 $$ 9. **Find corresponding $x$ values using $x = y + 1$:** - When $y=2$, $$ x = 2 + 1 = 3 $$ - When $y=-3$, $$ x = -3 + 1 = -2 $$ 10. **Final answer:** The points of intersection are: $$ (3, 2) \quad \text{and} \quad (-2, -3) $$