1. **State the problem:** Find the equation of a circle that touches the x-axis at the point $(3,0)$ and passes through the point $(1,2)$. 2. **Understand the problem:** A circle touching the x-axis at $(3,0)$ means the x-axis is tangent to the circle at that point. The tangent point lies on the circle, so $(3,0)$ is on the circle. 3. **Formula and properties:** The general equation of a circle is $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $(h,k)$ is the center and $r$ is the radius. 4. Since the circle touches the x-axis at $(3,0)$, the radius is the vertical distance from the center to the x-axis. The x-axis is $y=0$, so the radius $r = |k|$. 5. The tangent point $(3,0)$ lies on the circle, so: $$ (3 - h)^2 + (0 - k)^2 = r^2 = k^2 $$ 6. Simplify: $$ (3 - h)^2 + k^2 = k^2 $$ $$ (3 - h)^2 = 0 $$ $$ 3 - h = 0 $$ $$ h = 3 $$ 7. So the center is at $(3,k)$ and radius $r = |k|$. 8. The circle passes through $(1,2)$, so: $$ (1 - 3)^2 + (2 - k)^2 = k^2 $$ $$ (-2)^2 + (2 - k)^2 = k^2 $$ $$ 4 + (2 - k)^2 = k^2 $$ 9. Expand $(2 - k)^2$: $$ 4 + (4 - 4k + k^2) = k^2 $$ $$ 4 + 4 - 4k + k^2 = k^2 $$ $$ 8 - 4k + k^2 = k^2 $$ 10. Cancel $k^2$ on both sides: $$ 8 - 4k + \cancel{k^2} = \cancel{k^2} $$ $$ 8 - 4k = 0 $$ 11. Solve for $k$: $$ -4k = -8 $$ $$ k = 2 $$ 12. The center is $(3,2)$ and radius $r = |2| = 2$. 13. Write the equation of the circle: $$ (x - 3)^2 + (y - 2)^2 = 2^2 $$ $$ (x - 3)^2 + (y - 2)^2 = 4 $$ **Final answer:** $$ (x - 3)^2 + (y - 2)^2 = 4 $$