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Is line Y=KX + B tangent to circle X^2 + Y^2 = 1 ? 1. k+b=1

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Is line Y=KX + B tangent to circle X^2 + Y^2 = 1 ? 1. k+b=1 [#permalink]

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02 Mar 2008, 08:57
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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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02 Mar 2008, 12:28
Can't believe I spent so much time on this... I am going with E

I couldn't recollect any techniques or tips for tangents to circle, so I went ahead and substituted y=Kx+B in X^2+Y^2=1 in hope to find an intersection point. Then I got X^2*(1+K^2) + 2KBx + B^2 -1 = 0. When tried to see if the roots are real by trying to evaluate b^2 - 4ac, I got to 4K^2-4*B^2+4. And from statement 1 and 2 I couldn't figure if I get a real number or not. So picked E.

I am sure there must be an easy to do this, hope to see some replies here...
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02 Mar 2008, 22:54
sreehari wrote:
Can't believe I spent so much time on this... I am going with E

I couldn't recollect any techniques or tips for tangents to circle, so I went ahead and substituted y=Kx+B in X^2+Y^2=1 in hope to find an intersection point. Then I got X^2*(1+K^2) + 2KBx + B^2 -1 = 0. When tried to see if the roots are real by trying to evaluate b^2 - 4ac, I got to 4K^2-4*B^2+4. And from statement 1 and 2 I couldn't figure if I get a real number or not. So picked E.

I am sure there must be an easy to do this, hope to see some replies here...

it is very time-consuming though.

from 1., in case k=1 and b=0 we would have y=x which passes through the origin and is not tangent to the circle.
2. in case k=0 b could be 1 and thus y=1 which touches the circle. anyway it is only one point, since for k=1 and b=0 we have the same situation as in 1.

need quicker ways
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03 Mar 2008, 09:18
Ok.. found something that could be useful

formula to calculate pependicular distance from a point, say (p,q), to a line defined by ax+by+c=0 is

d = (ap+bq+c)/sqrt(a^2+b^2)

plug in (0,0) for (p,q) and 1 for d => B^2-K^2=1

Both statement 1 and 2 cannot be infered from the above equation. So still sticking to E.
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17 Mar 2008, 08:34
E

$$X^2 + Y^2 = 1$$ - is a circle that intersects Y-axis in points (0,1), (0,-1) and intersects X-axis in points (1,0), (-1,0).

Consider two edge examples that satisfy both conditions:
k=1; b=0 - y=x does not tangent to circle (intersects it)
k=0; b=1; - y=1 tangents to circle in the point (0,1).

Therefore, E
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17 Mar 2008, 08:42
Walker , I have come under the assumption that intersect can also be tangent. Is that correct?
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17 Mar 2008, 09:49
terp26 wrote:
Walker , I have come under the assumption that intersect can also be tangent. Is that correct?

http://en.wikipedia.org/wiki/Circle
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200px-CIRCLE_LINES_2.svg.png [ 14.95 KiB | Viewed 1055 times ]

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31 Mar 2008, 16:39
I agree its E ..

If either K or B =0

if k=1 then its not tangent

if b=1 then it is tangent..
Re: Coordinate geometry   [#permalink] 31 Mar 2008, 16:39
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