Known circle O:x? +y？ =4 and circle C:x? +(y-4)？ = 1; The center of the circle C moves the straight line M, and it intersects with the circle O at two points A and B. Question: Is there such a circle P in all the circles with diameter AB that the circle P passes through the point M (2,0)? If it exists, find the equation of circle p; If it does not exist, please explain the reason and find the slope of AB.

Solution: There are two such circles:

(1). When the straight line M passing through C (0 0,4) passes through the center O of the circle O, the intersection point AB with the circle O is the diameter of the circle O, and the circle with the diameter AB is the circle O. Of course, it passes through M (2 2,0) because M is on the circle O. ..

(2) Connect CM, then the intersection of the straight line M and the circle O where CM is located is A and M, and this M is the topic B, then the circle with the diameter of AM naturally intersects M. At this time, the slope k of the straight line M is -2, and the equation of M is y =-2 (x-2) =-2x+4; Substituting into the equation of circle o gives:

x？ +(-2x+4)？ =5x？ -16x+ 16=4, which is 5x? -16x+12 = (5x-6) (x-2) = 0, then x? =6/5，x？ =2; According to this, y? =- 12/5+4=8/5; y？ =0; That is, A(6/5, 8/5); B(2，0)。

∣AB∣=√(2-6/5)？ +(8/5)? ]=√(80/25)=√( 16/5)=4√( 1/5)

The coordinate of point P in AB: XP = (6/5+2)/2 = 8/5; Yp=(8/5)/2=4/5, then the equation of circle P is: (x-8/5)? +(y-4/5)？ =4/5.