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07 Apr 2020, 06:27
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Difficulty:
95%
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Question Stats:
33% (02:07) correct
67%
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wrong
based on 60
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Is line \(y=kx+b\) tangent to circle \(x^2+y^2=1\)?
(1) \(k+b=1\)
(2) \(b^2-k^2=1\)
(1) \(k+b=1\)
(2) \(b^2-k^2=1\)
07 Apr 2020, 15:37
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If \(y=kx+b\) is tangent to the circle, there must be a unique solution of both equations.
\(x^2 + (kx+b)^2 =1\)
\(x^2 + (kx)^2 +b^2 +2kxb -1 = 0\)
\((1+k^2) x^2 + 2kbx + b^2-1 = 0\)
If there is unique solution of this equation, discriminant of this equation is equal to 0.
\((2kb)^2 - 4 (b^2-1)(k^2+1) = 0\)
\(4k^2b^2 - 4k^2b^2 -4b^2 + 4k^2 +4 = 0\)
\(b^2 - k^2 = 1\)
Our Question stem is whether b^2 - k^2 = 1 or (b+k)(b-k) = 1.
Statement 1-
We have no idea about (b-k).
Insufficient
Statement 2-
\(b^2-k^2=1\)
That's exactly what we looking for.
Sufficient
\(x^2 + (kx+b)^2 =1\)
\(x^2 + (kx)^2 +b^2 +2kxb -1 = 0\)
\((1+k^2) x^2 + 2kbx + b^2-1 = 0\)
If there is unique solution of this equation, discriminant of this equation is equal to 0.
\((2kb)^2 - 4 (b^2-1)(k^2+1) = 0\)
\(4k^2b^2 - 4k^2b^2 -4b^2 + 4k^2 +4 = 0\)
\(b^2 - k^2 = 1\)
Our Question stem is whether b^2 - k^2 = 1 or (b+k)(b-k) = 1.
Statement 1-
We have no idea about (b-k).
Insufficient
Statement 2-
\(b^2-k^2=1\)
That's exactly what we looking for.
Sufficient
smyarga
prakhar992
23 Apr 2020, 15:19
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nick1816
Why the discriminant has to be Zero? Can you please explain? Is possible, via graph?
