If you consider both statements together:
k+b = 1; b=1-k; b^2 = (1-k)^2 = 1+k^2-2k
k^2+b^2 = 1; b^2= 1-k^2
Therefore, 1+k^2-2k = 1-k^2
On solving we get,
If k=1, b=0
Substituting these values in the equation y= kx+b, we get y=x.
If y=0, x=0.
Therefore, the line passes through the origin and is not a tangent.
Ans should be C
you forgot the other solution when k=0 b=1
y=1.. which is tangent to the circle.. (See Walker post above)
so two solutions... insuffient.
Your attitude determines your altitude
Smiling wins more friends than frowning