The equation (x-1)^2+(y-1)^2=1 is the equation of a circle with center at (1,1) and radius 1.
This simplifies to x^2-2x+1+y^2-2y+1=1 or x^2+y^2-2x-2y+1=0
As the line is y=mx, let us put this value of y in the equation of the circle and we get,
x^2+m^2x^2-2x-2mx+1=0 or (1+m^2)x^2-2(1+m)x+1=0
Hence using quadratic formula
x=(-(-2(1+m))+-sqrt((-2(1+m))^2-4(1+m^2)))/(2(1+m^2) or
x=((2(1+m))+-sqrt(4(1+m)^2-4(1+m^2)))/(2(1+m^2) or
x=((2(1+m))+-sqrt(4+4m^2+8m-4-4m^2))/(2(1+m^2) or
x=(2+2m+-sqrt(8m))/(2(1+m^2)
Now if m=0, we have x=1 and y=0 and hence, we have a tangent.
If m<0, there is no solution and line y=mx does not intersect circle.
If m>0, the let the two points will be D(x_1,mx_1) and E(x_2,mx_2) and distance between these points DE is
sqrt((x_2-x_1)^2+(mx_2-mx_1)^2)=|x_2-x_1|sqrt(1+m^2).
As x=(2+2m+-sqrt(8m))/(2(1+m^2), |x_2-x_1|=(2sqrt(8m))/(2(1+m^2))=sqrt(8m)/(1+m^2)
Hence DE=sqrt(8m)/(1+m^2)xxsqrt(1+m^2)=sqrt((8m)/(1+m^2))
