The straight line passing through the origin intersects with the circle X 2+Y 2-6x+5 = 0 at points A and B, and the locus equation of the middle point M of the chord AB is found.

Solution: It is known that x 2+y 2-6x+5 = 0.

- (X-3)^2+Y^2=4

This circle: the coordinate of point C is (3,0) and the radius is 2.

-Let the coordinates of m be (a, b).

It is known that m is the midpoint of AB, and m is on a straight line passing through the origin.

-Connect CM, then CM is perpendicular to AB, that is, CM is perpendicular to MO.

-In a right triangle, mo 2+cm 2 = oc 2.

-(a^2+b^2)+[(a-3)^2+(b-0)^2]=3^2

- (a-3/2)^2+b^2=9/4

The locus equation of the point m in the chord AB is (a-3/2) 2+b 2 = 9/4.