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M17-12

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M17-12  [#permalink]

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New post 16 Sep 2014, 01:01
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

51% (01:37) correct 49% (01:22) wrong based on 65 sessions

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Re M17-12  [#permalink]

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New post 16 Sep 2014, 01:01
2
1
Official Solution:


Notice that a circle represented by the equation \(x^2+y^2=1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

(1) \(k+b=1\). If \(k=0\) and \(b=1\) then the equation of the line becomes \(y=1\) and this line is tangent to the circle but if \(k=1\) and \(b=0\) then th equation of the line becomes \(y=x\) and this line is NOT tangent to the circle. Not sufficient.

(2) \(k^2+b^2=1\). The same example is valid for this statement too. Not sufficient.

(1)+(2) Again the same example satisfies both statement: if \(k=0\) and \(b=1\) then the equation of the line becomes \(y=1\) and this line is tangent to the circle but if \(k=1\) and \(b=0\) then th equation of the line becomes \(y=x\) and this line is NOT tangent to the circle. Not sufficient. Look at the diagram below to see both cases:

Image


Answer: E
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Re M17-12  [#permalink]

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New post 11 Jul 2016, 07:18
I think this is a high-quality question and I agree with explanation.
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Re: M17-12  [#permalink]

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New post 08 Mar 2017, 10:09
1
Here, x-intercept by putting y=0 is x=-b/k, if k=0 then x=not -defined in that case k cannot be zero hence, b=0 & k=1
Please let me know where i am going wrong?
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Re: M17-12  [#permalink]

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New post 26 Apr 2017, 07:53
The equation of the line can be written as y-kx-b=0. The distance of the line from center of the circle (origin in this case) is given by mod(b/sqrt(1+k^2)). (Distance of a line ax+by+c from x1,y1 is given by ax1+by1+c/sqrt(a^2+b^2) ).

For this line to be a tangent to the given circle , distance from origin to the line must be one.

Solving the two equations , we get b+sqrt(1+k^2) = 1 ; i,e, b^2-k^2 = 1.

Statement 1 : Not sufficient.

Statement 2 : Not sufficient.

Both the statements together : We get bk=0. b^2-k^2=1 or -1. Hence Not sufficient

Hence the answer must be E.
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Re: M17-12  [#permalink]

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New post 10 Jun 2018, 07:01
Please tell me where i went wrong.

If L; y=kx+b is a tangent to the circle then b2=k2+1.
(1) k+b=1=> b=1-k=> b2=(1-k)2=>1+k2= 1+k2-2k=>k=0, b=1 SUFF
(2)b2+k2=1=> k2+1+k2=1=>2k2=0=>k=0 SUFF

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Re: M17-12 &nbs [#permalink] 10 Jun 2018, 07:01
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