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09 Oct 2010, 01:53
Kudos
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Is line Y = KX + B tangent to circle \(X^2 + Y^2 = 1\) ?
1. K + B = 1
2. \(K^2 + B^2 = 1\)
A) Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
B) Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D) EACH statement ALONE is sufficient
E) Statements (1) and (2) TOGETHER are NOT sufficient
1. K + B = 1
2. \(K^2 + B^2 = 1\)
A) Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
B) Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D) EACH statement ALONE is sufficient
E) Statements (1) and (2) TOGETHER are NOT sufficient
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09 Oct 2010, 02:05
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Orange08
Lets solve the equations simultaneously. If a line is a tangent to the circle the equations will have exactly one solution which is the point of intersection.
\(y=kx+b\)
\(x^2+y^2=1\)
\(x^2+(kx+b)^2=1\)
\((1+k^2)x^2+2kbx+b^2-1=0\)
For a quadratic equation to have a single solution, the discriminant (b^2-4ac) must be 0. Therefore :
\((2kb)^2 - 4(1+k^2)(b^2-1)=0\)
\(4k^2b^2 - 4(b^2-1+k^2b^2-k^2)=0\)
\(-4b^2+4+4k^2=0\)
\(b^2-k^2=1\)
(1)k+b=1. Insufficient to show \(b^2-k^2=1\)
(2)\(b^2+k^2=1\), Insufficient to show \(b^2-k^2=1\)
(1+2) Both together are also insufficient
Answer is (e)
You can also refer to coordinate-geometry-69516.html for an example based approach
