x2+(y-2)2=2 is a circle with its center C at (0,2) and r=√(2) ( from (x-h)2+(y-k)2=r2 , std form of a circle )
y=mx meets this circle tangentially at point A in Q1, and point B in Q2. A & B are symmetrical about the y-axis because the circle is
centered on the y-axis, so lets find A, then B will be obvious. Also note y=mx passes through the origin O(0,0).
If you draw a triangle OAC you will find |OC|=2, |CA|=√(2) and, noting the tangent meets the circle at 90o, you can
solve for |OA|=√(2). Now a triangle drawn from A down to the x-axis and over to the origin is also a 45-45-90 triangle
and A(ax,ay) can be easily found (cos(45o) = √(2)/2, sin(45o) = √(2)/2)...
A is therefore at (1,1) and the value of m=1.
B is at (-1,1) and m=-1 for the other line.
There are probably more efficient methods to find m, A, and B. I just followed the first process that came to mind.