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

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Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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New post 07 Apr 2020, 05:27
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Question Stats:

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Is line \(y=kx+b\) tangent to circle \(x^2+y^2=1\)?

(1) \(k+b=1\)
(2) \(b^2-k^2=1\)
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Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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New post 07 Apr 2020, 14:37
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If \(y=kx+b\) is tangent to the circle, there must be a unique solution of both equations.

\(x^2 + (kx+b)^2 =1\)

\(x^2 + (kx)^2 +b^2 +2kxb -1 = 0\)

\((1+k^2) x^2 + 2kbx + b^2-1 = 0\)

If there is unique solution of this equation, discriminant of this equation is equal to 0.

\((2kb)^2 - 4 (b^2-1)(k^2+1) = 0\)

\(4k^2b^2 - 4k^2b^2 -4b^2 + 4k^2 +4 = 0\)

\(b^2 - k^2 = 1\)

Our Question stem is whether b^2 - k^2 = 1 or (b+k)(b-k) = 1.

Statement 1-

We have no idea about (b-k).

Insufficient

Statement 2-

\(b^2-k^2=1\)
That's exactly what we looking for.

Sufficient






smyarga wrote:
Is line \(y=kx+b\) tangent to circle \(x^2+y^2=1\)?

(1) \(k+b=1\)
(2) \(b^2-k^2=1\)
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Re: Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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New post 23 Apr 2020, 14:19
nick1816 wrote:
If \(y=kx+b\) is tangent to the circle, there must be a unique solution of both equations.

\(x^2 + (kx+b)^2 =1\)

\(x^2 + (kx)^2 +b^2 +2kxb -1 = 0\)

\((1+k^2) x^2 + 2kbx + b^2-1 = 0\)

If there is unique solution of this equation, discriminant of this equation is equal to 0.

\((2kb)^2 - 4 (b^2-1)(k^2+1) = 0\)

\(4k^2b^2 - 4k^2b^2 -4b^2 + 4k^2 +4 = 0\)

\(b^2 - k^2 = 1\)

Our Question stem is whether b^2 - k^2 = 1 or (b+k)(b-k) = 1.

Statement 1-

We have no idea about (b-k).

Insufficient

Statement 2-

\(b^2-k^2=1\)
That's exactly what we looking for.

Sufficient






smyarga wrote:
Is line \(y=kx+b\) tangent to circle \(x^2+y^2=1\)?

(1) \(k+b=1\)
(2) \(b^2-k^2=1\)



Why the discriminant has to be Zero? Can you please explain? Is possible, via graph?
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Re: Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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New post 23 Apr 2020, 15:03
Since the line cuts the circle at only one point, roots of the \(x^2 + (kx+b)^2 =1\) must be equal. {complex roots of a quadratic equation always occur in pair and are complementary, hence if you have one real root, other must be real.}

Red line doesn't cut the circle. For this line, discriminant of the equation is negative. (equation doesn't have any real root)

Brown line cuts the circle at 2 distinct points. For this line, discriminant of the equation is positive. (equation has 2 distinct roots or 2 distinct values of x)

Blue line (line given in our question) cuts the circle at one distinct point. For this line, discriminant of the equation is zero. (equation has 2 equal roots or 1 value of x)

If you still have any doubt, you can ask.


prakhar992 wrote:
nick1816 wrote:
If \(y=kx+b\) is tangent to the circle, there must be a unique solution of both equations.

\(x^2 + (kx+b)^2 =1\)

\(x^2 + (kx)^2 +b^2 +2kxb -1 = 0\)

\((1+k^2) x^2 + 2kbx + b^2-1 = 0\)

If there is unique solution of this equation, discriminant of this equation is equal to 0.

\((2kb)^2 - 4 (b^2-1)(k^2+1) = 0\)

\(4k^2b^2 - 4k^2b^2 -4b^2 + 4k^2 +4 = 0\)

\(b^2 - k^2 = 1\)

Our Question stem is whether b^2 - k^2 = 1 or (b+k)(b-k) = 1.

Statement 1-

We have no idea about (b-k).

Insufficient

Statement 2-

\(b^2-k^2=1\)
That's exactly what we looking for.

Sufficient






smyarga wrote:
Is line \(y=kx+b\) tangent to circle \(x^2+y^2=1\)?

(1) \(k+b=1\)
(2) \(b^2-k^2=1\)



Why the discriminant has to be Zero? Can you please explain? Is possible, via graph?

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Re: Is line y=kx+b tangent to circle x^2+y^2=1?   [#permalink] 23 Apr 2020, 15:03

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

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