X 2+y 2+x-y-1= 0 = = > when m =-1; (x+ 1/2)^2+(y- 1/2)^2=3/2

Let the linear L equation be y = kx+1; M(x 1，y 1)，N(x2，y2)

If OM is perpendicular to ON, then (y1/x1) * (y2/x2) =-1means x 1x2+y 1y2=0.

Substitute y=kx+ 1 into x 2+y 2+x-y- 1 = 0.

Simplified as (1+k2) x2+(k+1) x-1= 0.

x 1+x2=-(k+ 1)/( 1+k^2)； x 1x2=- 1/( 1+k^2)

y 1+y2=kx 1+kx2+2=k(x 1+x2)+2=(k^2-k+2)/(k^2+ 1)

y 1y2=(kx 1+ 1)(kx2+ 1)= 1+k(x 1+x2)+x 1x2k^2=(-k^2-k+ 1)/(k^2+ 1)

x 1x2+y 1y2=0

That is-1(1+k2)+(-k2-k+1)/(k2+1) = 0.

K 1=0，k2=- 1。

So the linear equation is y= 1 or x+y- 1=0.