KASSALMD wrote:

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,

k=1

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.

E.

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