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# Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 1. k +

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Joined: 11 May 2008
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Is line y = kx + b tangent to circle x^2 + y^2 = 1 ? 1. k + [#permalink]

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30 Aug 2008, 03:34
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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30 Aug 2008, 03:47
E

Consider two simple examples that satisfy both conditions:

a) k=1, b=0: y=x - goes thorough (0,0) point and intersect the circle at two points. false.
b) k=0, b=1: y=1 - is a tangent of the circle. true.
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30 Aug 2008, 04:45
.. whenever i see ur bike...
i dont know which is faster...
ur mind or the bike....

take a bow serg!!!
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30 Aug 2008, 04:54
I've just finished my draft of one of HBS essays and decided to relax a bit at math forum

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30 Aug 2008, 05:23
walker wrote:
I've just finished my draft of one of HBS essays and decided to relax a bit at math forum

Walker = Ultimate Math Machine
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03 Sep 2008, 08:27
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
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03 Sep 2008, 08:32
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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03 Sep 2008, 09:48
Manager
Joined: 03 Jun 2008
Posts: 128
Schools: ISB, Tuck, Michigan (Ross), Darden, MBS

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04 Sep 2008, 04:03
arjtryarjtry wrote:
Is line $$y = kx + b$$ tangent to circle $$x^2 + y^2 = 1$$ ?

1. $$k + b = 1$$
2. $$k^2 + b^2 = 1$$

nice explanation Mr. Walker!

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Re: coordinate geometry   [#permalink] 04 Sep 2008, 04:03
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