1. **State the problem:** Find the points of intersection between the circle given by the equation $$ (x - 5)^2 + (y - 2)^2 = 36 $$ and the vertical line $$ x = 5 $$. 2. **Use the formula:** The circle equation represents all points $$ (x,y) $$ at a distance 6 (since $$ \sqrt{36} = 6 $$) from the center $$ (5,2) $$. The line $$ x=5 $$ is vertical through the center. 3. **Substitute $$ x=5 $$ into the circle equation:** $$ (5 - 5)^2 + (y - 2)^2 = 36 $$ $$ 0 + (y - 2)^2 = 36 $$ $$ (y - 2)^2 = 36 $$ 4. **Solve for $$ y $$:** $$ y - 2 = \pm 6 $$ So, $$ y = 2 + 6 = 8 $$ $$ y = 2 - 6 = -4 $$ 5. **Final points of intersection:** $$ (5, 8) \quad \text{and} \quad (5, -4) $$ --- 1. **State the problem:** Find the points of intersection between the circle $$ (x - 5)^2 + (y - 2)^2 = 36 $$ and the horizontal line $$ y = 8 $$. 2. **Substitute $$ y=8 $$ into the circle equation:** $$ (x - 5)^2 + (8 - 2)^2 = 36 $$ $$ (x - 5)^2 + 6^2 = 36 $$ $$ (x - 5)^2 + 36 = 36 $$ 3. **Simplify:** $$ (x - 5)^2 = 0 $$ 4. **Solve for $$ x $$:** $$ x - 5 = 0 $$ $$ x = 5 $$ 5. **Final point of intersection:** $$ (5, 8) $$ **Summary:** The vertical line $$ x=5 $$ intersects the circle at two points: $$ (5,8) $$ and $$ (5,-4) $$. The horizontal line $$ y=8 $$ intersects the circle at exactly one point: $$ (5,8) $$.