sreehari wrote:

Can't believe I spent so much time on this... I am going with E

I couldn't recollect any techniques or tips for tangents to circle, so I went ahead and substituted y=Kx+B in X^2+Y^2=1 in hope to find an intersection point. Then I got X^2*(1+K^2) + 2KBx + B^2 -1 = 0. When tried to see if the roots are real by trying to evaluate b^2 - 4ac, I got to 4K^2-4*B^2+4. And from statement 1 and 2 I couldn't figure if I get a real number or not. So picked E.

I am sure there must be an easy to do this, hope to see some replies here...

it is very time-consuming though.

from 1., in case k=1 and b=0 we would have y=x which passes through the origin and is not tangent to the circle.

2. in case k=0 b could be 1 and thus y=1 which touches the circle. anyway it is only one point, since for k=1 and b=0 we have the same situation as in 1.

need quicker ways