1. **State the problem:** Find all points $(x,y)$ such that $y=3$ and the distance from $(-2,5)$ to $(x,3)$ is 8 units. 2. **Formula used:** The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Apply the formula:** Here, $d=8$, $(x_1,y_1)=(-2,5)$, and $(x_2,y_2)=(x,3)$. $$8 = \sqrt{(x + 2)^2 + (3 - 5)^2}$$ 4. **Simplify inside the square root:** $$(3 - 5)^2 = (-2)^2 = 4$$ 5. **Square both sides to remove the square root:** $$64 = (x + 2)^2 + 4$$ 6. **Isolate $(x + 2)^2$:** $$(x + 2)^2 = 64 - 4 = 60$$ 7. **Solve for $x + 2$:** $$x + 2 = \pm \sqrt{60} = \pm 2\sqrt{15}$$ 8. **Find $x$ values:** $$x = -2 \pm 2\sqrt{15}$$ 9. **Final points:** $$(x,y) = \left(-2 + 2\sqrt{15}, 3\right) \text{ and } \left(-2 - 2\sqrt{15}, 3\right)$$ These are the two points with $y=3$ that are 8 units away from $(-2,5)$.