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

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Tutor
Joined: 20 Apr 2012
Posts: 98
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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07 Apr 2020, 05:27
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Difficulty:

85% (hard)

Question Stats:

30% (01:49) correct 70% (02:15) wrong based on 30 sessions

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

(1) $$k+b=1$$
(2) $$b^2-k^2=1$$
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 1890
Location: India
Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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07 Apr 2020, 14:37
2
1
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) $$b^2-k^2=1$$
Intern
Joined: 05 May 2019
Posts: 48
Re: Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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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?
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 1890
Location: India
Re: Is line y=kx+b tangent to circle x^2+y^2=1?  [#permalink]

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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