If two circles are circumscribed, then the inner common tangent bisects the outer common tangent (the line segment between two tangent points) of the two circles How to prove

If two circles are circumscribed, then the inner common tangent bisects the outer common tangent (the line segment between two tangent points) of the two circles How to prove

Let p be the circumtangent point of the two circles, a and B be the circumtangent point of the two circles, and a and B be the intersection point of the inner common tangent point of the two circles
According to the tangent length theorem, AC = CP, BC = CP, the inner common tangent bisects the outer common tangent of two circles

If the moving circle with radius 1 is known to be tangent to the circle (X-5) 2 + (y + 7) 2 = 16, then the trajectory equation of the center of the moving circle is______ ．

The center of the circle (X-5) 2 + (y + 7) 2 = 16 is C (5, - 7), and the radius is r = 4. ∵ the moving circle with radius 1 is tangent to the circle (X-5) 2 + (y + 7) 2 = 16, | when the two circles are inscribed, the distance from the center a of the moving circle to the point C is equal to the absolute value of the difference between the radii of the two circles, | BC | = 4-1 = 3, so the trajectory of the center of the moving circle is a circle with radius 3, and the trajectory equation is (X-5) 2 + (y + 7) 2 = 9; When two circles are circumscribed, the distance from the center B of the moving circle to the point C is equal to the sum of the radii of the two circles, | BC | = 4 + 1 = 5, so the trajectory of the center of the moving circle is a circle with C as the center and radius equal to 5, and the trajectory equation is (X-5) 2 + (y + 7) 2 = 25. To sum up, the trajectory equation of the center of the moving circle is (X-5) 2 + (y + 7) 2 = 9 or (X-5) 2 + (y + 7) 2 = 25 （x-5）2+（y+7）2=25．