(2x+y-7)m+x+y-4=0

Because 2x+y-7=0 and x+y-4=0 intersect (3, 1).

Therefore, the straight line (2m+1) x+(m+1) y-7m-4 = 0.

Constant intersection (3, 1)

Circle c: (x- 1) square +(y-2) square =25 center (1, 2), radius R=5.

Distance from center c to fixed point (3, 1) d:

d=√[(3- 1)^2+( 1-2)^2]=√5

Because: R>D

So the fixed point (3, 1) on the straight line L is on the circle C.

So: No matter what real number M takes, the straight line L intersects the circle and intersects two points.

Center (1, 2), radius r=5.

Center distance = | (2m+1)+2 (m+1)-7m-4 |/√ [(2m+1)+(m+1) 2]

=|3m+ 1|/√(5m^2+6m+2)

Then (half the chord length) 2 = R2- center distance 2.

So it is to find the maximum of the center distance of 2.

Center distance 2 = a = (3m+1) 2/(5m 2+6m+2)

=(9m^2+6m+ 1)/(5m^2+6m+2)

5am^2+6am+2a=9m^2+6m+ 1

(5a-9)m^2+(6a-6)m+(2a- 1)=0

The solution of this equation must be

(6a-6)^2-4(5a-9)(2a- 1)≥0

a^2-5a≤0

0≤a≤5

Therefore, the maximum distance between centers is =√5.

So at this time, half the chord length =2√5.

So the minimum chord length =4√5.

Substitute a=5 for (5a-9) m 2+(6a-6) m+(2a-1) = 0.

(4m+3)^2=0

m=-3/4