The tangent point of the inner common tangent is recorded as the triangle formed by two points c and PQ is a right triangle,

Therefore, the midpoint of PQ is marked as m, which is the same as the abscissa of the tangent point of the internal common tangent. Let M(x, y).

So cm = PQ, the radius of circle O 1 is x+3, and the radius of circle O2 is 3-X..

Let the radius O 1 of a circle be smaller than the radius O2 (that is, X.

Connect O 1P and O2Q, and make O 1 d perpendicular to O2Q through O1to form a right triangle.

According to Pythagorean theorem, 4Y 2+4x 2 = 36, that is, X 2+Y 2 = 9.

In fact, when x>0, the trajectory equation is the same.

To sum up, the trajectory equation of PQ midpoint is x 2+y 2 = 9 (-3