l= sqrt (2^2 -1) =sqrt(3) =1.732.

Now draw a line perpendicular to the y axis to the point where the tangent cuts the circle. OK? The length of that line would be 1.732 sin(a), where “a” is the angle corresponding to sin(a) =0.5.

So your tangent cuts circle at [(1.732 sin(a) , 1-1.732 cos(a)]

and the second tangent will cut at [(-1.732sin(a),1-1.732cos(a)]

So your points are (0.866, -0.5) and (-0.866, -0.5)

You can also do this as a algebra problem. assume the equation of the tangent is y = cx +1 [ this passes through (0,1)]. Substitute this in your circle equation.

You get (c^2 +1)x^2 +c^2(x^2) +3 = 0 . Then you find the 2 roots of this qudratic equation as

x1 , x2 = {-4c +/- sqrt [16c^2 – 12(c^2 +1)]}/ (c^2+1)

Substitute in y = cx +1 and you get y1 and y2

So your points of intersection will be (x1,y1) and( x2,y2)

Substitute these points in the circle equation y = cx + 1 and you can evaluate c in terms of x1 and y1. It may get little tedious, but you can do it by this methos also.