1. **State the problem:** We need to find the coordinates $(x,y)$ such that the distance from the point $(x,y)$ to the point $(-1,5)$ is $\sqrt{2}$. This is given by the equation $$\sqrt{2} = (x+1)^2 + (y-5)^2.$$ 2. **Recall the distance formula:** The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$ 3. **Rewrite the problem:** The equation given is $$\sqrt{2} = (x+1)^2 + (y-5)^2,$$ but this is incorrect because the right side should be under the square root. The correct distance formula is $$\sqrt{2} = \sqrt{(x+1)^2 + (y-5)^2}.$$ 4. **Square both sides to eliminate the square root:** $$2 = (x+1)^2 + (y-5)^2.$$ 5. **Interpretation:** This equation represents a circle centered at $(-1,5)$ with radius $\sqrt{2}$. The points $(x,y)$ lie on this circle. 6. **Express $y$ in terms of $x$ or vice versa:** $$ (y-5)^2 = 2 - (x+1)^2. $$ 7. **Solve for $y$:** $$ y - 5 = \pm \sqrt{2 - (x+1)^2}, $$ $$ y = 5 \pm \sqrt{2 - (x+1)^2}. $$ 8. **Domain restriction:** For $y$ to be real, the expression under the square root must be non-negative: $$ 2 - (x+1)^2 \geq 0 \implies (x+1)^2 \leq 2. $$ 9. **Final solution:** The coordinates $(x,y)$ satisfy $$ y = 5 \pm \sqrt{2 - (x+1)^2} $$ with $$ -1 - \sqrt{2} \leq x \leq -1 + \sqrt{2}. $$ This describes all points on the circle with center $(-1,5)$ and radius $\sqrt{2}$.