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# M17-12

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Math Expert
Joined: 02 Sep 2009
Posts: 52437

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16 Sep 2014, 00:01
00:00

Difficulty:

85% (hard)

Question Stats:

50% (01:37) correct 50% (01:21) wrong based on 66 sessions

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Is line $$y = kx + b$$ tangent to circle $$x^2 + y^2 = 1$$ ?

(1) $$k + b = 1$$

(2) $$k^2 + b^2 = 1$$

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Joined: 02 Sep 2009
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16 Sep 2014, 00:01
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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:

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11 Jul 2016, 06:18
I think this is a high-quality question and I agree with explanation.
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Joined: 01 Sep 2014
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08 Mar 2017, 09:09
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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26 Apr 2017, 06:53
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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10 Jun 2018, 06:01
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

Ans (D)
Re: M17-12 &nbs [#permalink] 10 Jun 2018, 06:01
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# M17-12

Moderators: chetan2u, Bunuel

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