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Geometry... least possible distance

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by eagleeye » Mon Jul 30, 2012 7:59 pm
topspin360 wrote:can anyone show how to do this one?

thanks.

What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = (3/4)x − 3 ?

1.4
sqrt(2)
1.7
sqrt(3)
2.0
See the sketch attached. It is pretty self-explanatory except for a couple of things.
For the given line equation, we have, the x and y intercepts as
y=(3/4)x-3

When x=0, y=-3,
When y=0, x=4.

Then draw the perpendicular from the origin to the line segment joining A and B. Call this point C. We need to find OC.

We have the equation of the circle as x^2+y^2 = 1.
Before we go any further, here's a disclaimer. We can't be expected to solve this one in under 2 minutes, since we have no idea (as far as GMAT is concerned) what x^2+y^2 = 1 signifies.

For the purpose of solving this question, I am going to tell you what it means. It tells us that the circle is centered at (0,0) and has a radius of 1. Since we are not expected to know this, this question as written would not be a valid GMAT question. Anyway, since I do know what it means, I have solved the question anyway, assuming that we were told in the question stem itself that the circle is centered at (0,0) and has a radius of 1. With that in mind, and the attached figure in sight, let's continue.

Now we need to find CD. If we can find OC, we will find CD = OC-OD = OC - 1. Rest of the solution follows in the attached figure.

Let me know if this helps :)
Attachments
Mindistance.jpg
Last edited by eagleeye on Mon Jul 30, 2012 8:56 pm, edited 3 times in total.
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by alex.gellatly » Mon Jul 30, 2012 8:12 pm
eagleeye wrote:
Key takeaways from this question:

Equation of a standard circle centered at (a,b) with radius r is:
(x-a)^2 + (y-b)^2 = r^2

Distance of a line with equation ax+by+c=0 from a point (m,n) is:

|am+bn+c|/sqrt(a^2+b^2)

Let me know if this helps :)
I haven't seen these equations nor these types of problems in my GMAT prep thus far. Should they be known or is there another approach? Thanks
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
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by eagleeye » Mon Jul 30, 2012 8:37 pm
alex.gellatly wrote:
eagleeye wrote:
Key takeaways from this question:

Equation of a standard circle centered at (a,b) with radius r is:
(x-a)^2 + (y-b)^2 = r^2

Distance of a line with equation ax+by+c=0 from a point (m,n) is:

|am+bn+c|/sqrt(a^2+b^2)

Let me know if this helps :)
I haven't seen these equations nor these types of problems in my GMAT prep thus far. Should they be known or is there another approach? Thanks
Good point. I started working on the alternative as soon as I came out of my Math Trance. See the GMAT version in the original post :). You need to know what x^2+y^2 = 1 means to solve the question.

This equation is not covered in the GMAT math review under circles or co-ordinate geometry. Hence I would imagine, this exact question would never appear on the GMAT itself.

The best version that would work will say,

If a circle is centered at (0,0) and has radius of 1, what is the shortest distance between the circle and the line with the equation y=3x/4-3.

As the question is written right now, it is next to impossible to solve within the stipulated 2 minutes, unless you know what x^2+y^2 = 1 means, which as a GMAT person, you are not supposed to know.

Anyway, the solution above shows a couple of important points, which make this problem a helpful one.

Let me know if this helps :)
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by GMATGuruNY » Tue Jul 31, 2012 11:31 am
topspin360 wrote: What is the least possible distance between a point on the line y = (3/4)x − 3 and a point on the circle centered at the origin with a radius of 1?

1.4
sqrt(2)
1.7
sqrt(3)
2.0
I've changed the wording of the problem so that the solution does not require knowledge of the equation of a circle.

DRAW the figure:
Image

The SHORTEST DISTANCE between a point and a line must form a RIGHT ANGLE with the line.
Thus:
AC = the shortest distance between the center of the circle and y = (3/4)x - 3.
EC = the shortest distance between a point on the circle and y = (3/4)x - 3.

∆ABD is a 3-4-5 triangle.
Area of ∆ABD = (1/2)(AD)(AB) = (1/2)(3)(4) = 6.
If we consider BD the base of ∆ABD, then AC is the corresponding height.
Since 1/2 the product of any base and height must yield the same area, we get:
(1/2)(BD)(AC) = 6
(1/2)(5)(AC) = 6.
AC = 12/5.

As the radius of the circle, AE = 1.
Thus, EC = AC - AE = 12/5 - 1 = 7/5 = 1.4.

The correct answer is A.
Last edited by GMATGuruNY on Tue Jul 31, 2012 6:38 pm, edited 1 time in total.
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by topspin360 » Tue Jul 31, 2012 6:29 pm
thanks all for each amazing explanations! really appreciate your effort in helping out. this problems came from one of the GMATclub tests... which are supposedly targeted at 750+ problems. But i guess some problems just go way too far.
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