1. **State the problem:** Solve the simultaneous equations: $$x^2 + y^2 = 13$$ $$x = y - 5$$ 2. **Substitute the linear equation into the circle equation:** Since $x = y - 5$, substitute $x$ in the first equation: $$ (y - 5)^2 + y^2 = 13 $$ 3. **Expand and simplify:** $$ (y - 5)^2 = y^2 - 10y + 25 $$ So, $$ y^2 - 10y + 25 + y^2 = 13 $$ Combine like terms: $$ 2y^2 - 10y + 25 = 13 $$ 4. **Bring all terms to one side:** $$ 2y^2 - 10y + 25 - 13 = 0 $$ $$ 2y^2 - 10y + 12 = 0 $$ Divide entire equation by 2: $$ y^2 - 5y + 6 = 0 $$ 5. **Factor the quadratic:** $$ (y - 2)(y - 3) = 0 $$ So, $$ y = 2 \quad \text{or} \quad y = 3 $$ 6. **Find corresponding $x$ values using $x = y - 5$:** - For $y = 2$: $$ x = 2 - 5 = -3 $$ - For $y = 3$: $$ x = 3 - 5 = -2 $$ 7. **Final solutions:** $$ (-3, 2) \quad \text{and} \quad (-2, 3) $$ These points satisfy both equations simultaneously.