**Question Stem:** 如图，一动圆与圆 $O_1: (x+3)^2 + y^2 = 1$ 外切，与圆 $O_2: (x-3)^2 + y^2 = 81$ 内切，求动圆圆心的轨迹方程. **Diagram Description:** * **Type:** Geometric diagram with coordinate axes. * **Elements:** * **Coordinate Axes:** X-axis and Y-axis intersecting at the origin O. * **Origin:** Point O labeled at (0,0). * **Circles:** * A smaller circle centered at $O_1$ on the negative X-axis. * A larger circle centered at $O_2$ on the positive X-axis. * A third circle, representing the "moving circle", centered at point P. This moving circle is tangent to both the smaller and larger circles. In the lower diagram, point P is shown within the larger circle but outside the smaller circle. The smaller circle is entirely inside the larger circle. * **Points:** $O_1$ (left of origin on X-axis), $O_2$ (right of origin on X-axis), P (center of the moving circle). * **Lines:** Lines are drawn from O to $O_1$, O to $O_2$. In the lower diagram, lines are drawn from P to $O_1$ and P to $O_2$. * **Labels:** Points O, $O_1$, $O_2$, P are labeled. The axes are labeled x and y. **Solution/Explanation Content:** 圆 $O_1: (x+3)^2 + y^2 = 1$ 的圆心为 $O_1(-3,0)$，半径 $r_1 = 1$. 圆 $O_2: (x-3)^2 + y^2 = 81$ 的圆心为 $O_2(3,0)$，半径 $r_2 = 9$. $|O_1 O_2| = 6 < r_2 - r_1$，所以圆 $O_1$ 与圆 $O_2$ 的关系是内含. 设动圆圆心为 $P(x, y)$，动圆半径为 $r$, 由于动圆与圆 $O_1$ 外切，所以 $|PO_1| = r + r_1 = r + 1$. 由于动圆与圆 $O_2$ 内切，所以 $|PO_2| = r_2 - r = 9 - r$. 因此，$|PO_1| + |PO_2| = (r+1) + (9-r) = 10$. Also, $|O_1 O_2| = \sqrt{(3 - (-3))^2 + (0-0)^2} = \sqrt{6^2} = 6$. So, $|PO_1| + |PO_2| = 10 > 6 = |O_1 O_2|$. 所以 $P$ 点的轨迹是以 $O_1, O_2$ 为焦点, 即 $c=3, a=5, b=4$ 的椭圆, 所以 $P$ 点的轨迹方程为 $\frac{x^2}{25} + \frac{y^2}{16} = 1$.