1. **Problem statement:** We have a circle $C_1$ with diameter endpoints $A(-2,18)$ and $B(14,6)$. We need to find the equation of $C_1$. Then, given another circle $C_2$ centered at the origin that touches $C_1$, find possible equations for $C_2$. 2. **Equation of circle from diameter:** The center of the circle is the midpoint of $AB$: $$\text{Center } M = \left(\frac{-2+14}{2}, \frac{18+6}{2}\right) = (6, 12)$$ 3. **Radius of $C_1$:** Radius $r$ is half the length of $AB$: $$AB = \sqrt{(14 - (-2))^2 + (6 - 18)^2} = \sqrt{16^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20$$ $$r = \frac{20}{2} = 10$$ 4. **Equation of $C_1$:** Using center-radius form: $$ (x - 6)^2 + (y - 12)^2 = 10^2 = 100 $$ 5. **Circle $C_2$ centered at origin:** Equation is: $$ x^2 + y^2 = R^2 $$ where $R$ is the radius of $C_2$. 6. **Condition for circles to touch:** Two circles touch externally if the distance between centers equals the sum of radii. Distance between centers: $$ d = \sqrt{(6 - 0)^2 + (12 - 0)^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5} $$ 7. **Touching condition:** $$ d = r_1 + r_2 $$ $$ 6\sqrt{5} = 10 + R $$ $$ R = 6\sqrt{5} - 10 $$ 8. **Alternatively, circles can touch internally:** $$ d = |r_1 - r_2| $$ $$ 6\sqrt{5} = |10 - R| $$ Two cases: - $10 - R = 6\sqrt{5} \Rightarrow R = 10 - 6\sqrt{5}$ - $R - 10 = 6\sqrt{5} \Rightarrow R = 10 + 6\sqrt{5}$ (not possible since $R$ must be positive and less than $d$ for internal touch) 9. **Possible radii for $C_2$:** $$ R = 6\sqrt{5} - 10 \quad \text{or} \quad R = 10 - 6\sqrt{5} $$ Note: $10 - 6\sqrt{5}$ is negative (approx -3.416), so discard. 10. **Final equations for $C_2$:** $$ x^2 + y^2 = (6\sqrt{5} - 10)^2 $$ **Summary:** - Equation of $C_1$: $$(x - 6)^2 + (y - 12)^2 = 100$$ - Equation of $C_2$: $x^2 + y^2 = (6\sqrt{5} - 10)^2$ This completes the solution.