Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: least distance [#permalink]
21 Nov 2008, 20:31

2

This post received KUDOS

This is a tough problem. I think most people are going to work on it for 2 min on test day and then make an educated guess.

The circle has 0,0 as the origin and a radius of 1. The line has y-intercept of -3 and runs up and to the right with a slope of +3/4. So it also has coordinates of (0,0), (0,-3) and (0,4). This means we have a 3-4-5 triangle.

The question asks for the least possible distance (a.k.a. the closes point) between the circle and the line given by the formula. The closest point of the circle to the line can be found by creating a second line. The slope of this line must be inverted and negated (slope of 3/4, becomes -4/3) and the y-intercept will be 0, meaning the line goes through the origin. Where this newly created line passes through the circle will be the closest point to the line. Now we just have to figure out the distance between these two points.

I first started off by drawing a picture. Sorry my photoshop isn't working or I'd upload a picture. But first draw the circle with radius 1 around the origin. I made 1 point at 0,-3 for the line's y-intercept and one dot at 4,0 for the x-intercept. I then drew a line as a hypotnuse of a triangle created by the x & y axis and the line. The hypotnuse I labeled c, the line from 0,0 to 0,-3 i labeled a and the line from 0,0 to 4,0 i labeled b. So we have triangle with sides a, b, and c.

Now the prependicular line created by y = -4/3x (inverted and negated slope of the line given) will divide this hypotnuse. I labeled the distance from the origin to the point where the original line and perpendicular cross as x. I also labeled the distance from (0,-3) to the perpendicular as d and the point from the perpendicular to (4,0) as e.

Because we know side a = 3, side b = 4, this means side c = 5

So we have a^2 + b^2 = c^2, d^2 + x^2 = a^2, and e^2 + x^2 = b^2 do some substitutions.

d^2 + x^2 = 9 & e^2 + x^2 = 16.

Because side c = d + e, we know that c = 5, so 5 = d + e. This can also be written d = 5 - e and e = 5 - d. These will become important later.

I've found it easier to combine 2 equations where possible. This can be done with both equations = the same thing, like x^2.

So our perpendicular created 2 smaller triangles within the larger one. Trianle ADX and EBX. Let's represent these by:

d^2 + x^2 = a^2 & e^2 + x^2 = b^2 Sub in the values we know for a and b d^2 + x^2 = 9 & e^2 + x^2 = 16 x^2 = 9 - d^2 and x^2 = 16 - e^2 Combine these;

x^2 + x^2 = 9 - d^2 + 16 - e^2; we have the variable d in there that we need to find a way to sub in to allow us to solve

d = 5 - e (from above) so d^2 = (5 - e)^2

2x^2 = 9 - (5-e)^2 + 16 - e^2; do FOIL on (5-e)^2

2x^2 = 9 - (25 - 10e + e^2) + 16 - e^2; now get rid of those ( ) by factoring the - into the terms

2x^2 = 9 -25 + 10e - e^2 + 16 - e^2. We see that we have e and e^2. Now what? well, we have values we can sub in for e^2 that include an x as a variable. This is good because we don't want to substitute in and keep changeing variables and never get done subbing in variables. So what does e^2 equal in terms of x? e^2 = 16 - x^2 so now sub this in

2x^2 = 9 - 25 + 10e - (16 - x^2) + 16 - (16 - x^2); The parenthesis are important because we're subbing and we don't want this to throw our values off. If the term subbed into the problem is + but () are not used, then the second term will not be correct in terms of + or -. Do some cleaning up to factor in and combine terms were possible.

2x^2 = -16 + 10e - 16 + x^2 + 16 - 16 + x^2; do some housecleaning and combine terms

2x^2 = 10e - 32 + 2x^2

32 = 10e

e = 3.2

remember that e is the distance from (4,0) to the perpendicular. This can now be used to find the value of x because e^2+x^2 = b^

Now we're not done (for reason #1, 2.4 doesn't match any answer!) But this is the distance from the perpendicular to the origin. But we're looking for the closest point in the circle to the line. What was the radius of the circle? 1 so we need to take 2.4 - 1 to find the distance of the point on the actual circle to the line.

Answer A.

There may be faster ways to answer this on test day, but if you get something like this, take a look to see if there is anything obvious, but then guess and move on. You're better off not wasting your time.

ritula wrote:

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

1.4 sqrt(2) 1.7 sqrt(3) 2

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: least distance [#permalink]
22 Nov 2008, 00:46

3

This post received KUDOS

yes, A jallenmorris, nice explanation, but I'm not sure this way won't take more than 2 min. I think the shortest way is as following:

We have right triangle with sides 3,4,5. Area is 1/2 (3x4) = 6. The shortest distance from the origin to the line y=3/4x-3 has to be perpendicular to this line and is 6/(5/2)=2,4. Thus, subtracting radius we have the answer: 2,4-1=1,4

Re: least distance [#permalink]
22 Nov 2008, 04:32

1

This post received KUDOS

Thats a very nice approach prasun. But i guess line should pass close to the circle on the 4th quadrant not 2nd quarter. wht do u say. But +1 to u for this approach.

prasun84 wrote:

A is correct.

the circle has origin at (0,0) and radius=1. the line y=3/4x-3 passes close to the circle on the 2nd quadrant....(we can plot the points and check..)

now the least distance between any point on the circle and line is the perpendicular to the line.

the length of the perpendicular from the centre(0,0)= 3/sqrt(1^2 +(3/4)^2) =2.4. Thus, distance from point on circle to line =2.4- 1 =1.4.

Re: least distance [#permalink]
22 Nov 2008, 05:26

the line meets Y-axis at -3 and x-axis at +4. That makes it 2nd quadrant. anyways, the solution by atleticos is better...if u can visualize the 3-4-5 triangle it can be solved almost in a minute.

ritula wrote:

Thats a very nice approach prasun. But i guess line should pass close to the circle on the 4th quadrant not 2nd quarter. wht do u say. But +1 to u for this approach.

prasun84 wrote:

A is correct.

the circle has origin at (0,0) and radius=1. the line y=3/4x-3 passes close to the circle on the 2nd quadrant....(we can plot the points and check..)

now the least distance between any point on the circle and line is the perpendicular to the line.

the length of the perpendicular from the centre(0,0)= 3/sqrt(1^2 +(3/4)^2) =2.4. Thus, distance from point on circle to line =2.4- 1 =1.4.

Re: least distance [#permalink]
22 Nov 2008, 05:50

I totally agree that my the way I went about the answer is long, but my point wasn't necessarily to lay out a road map for use on the GMAT. It would simply take too long. I've seen the forumula you guys used but it didn't make much sense to me at the time, so I went the backwards way to the answer to see how the forumula works.

The forumula is certainly faster. Can someone label the formula?

It also seems that this formula is kind of obscure for GMAT purposes. I don't recall GMAT authors writing questions that required remembering such formulas.

One last thing...I don't think my explanation would have been so long had I not been trying to eplain the rational behind each move. I believe it's important to show people how to think in approaching triangle questions. Someone may not get a question like this on their GMAT but they will have an easier time with triangles because they did this as a practice problem.

prasun84 wrote:

the line meets Y-axis at -3 and x-axis at +4. That makes it 2nd quadrant. anyways, the solution by atleticos is better...if u can visualize the 3-4-5 triangle it can be solved almost in a minute.

ritula wrote:

Thats a very nice approach prasun. But i guess line should pass close to the circle on the 4th quadrant not 2nd quarter. wht do u say. But +1 to u for this approach.

prasun84 wrote:

A is correct.

the circle has origin at (0,0) and radius=1. the line y=3/4x-3 passes close to the circle on the 2nd quadrant....(we can plot the points and check..)

now the least distance between any point on the circle and line is the perpendicular to the line.

the length of the perpendicular from the centre(0,0)= 3/sqrt(1^2 +(3/4)^2) =2.4. Thus, distance from point on circle to line =2.4- 1 =1.4.

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: least distance [#permalink]
22 Nov 2008, 07:51

agree with you....but i am of the opinion that knowing certain formulas can help you ace difficult questions in quick time...of course goes to say that it is essential to know how those formulas are derived....many a time, i have spent considerable time in trying to derive certain formulas which is an intermittent step in solving the real question....knowing those formulas sometimes has great advantage....esp under time constraints... say for example, a travels 30km at 30 km/hr and another 30 km at 80km/hr.what is the average speed? i could calculate the time taken and then try divide total distance/total time. however, if equal distances are covered by unequal speeds, my average speed is the harmonic mean of the two speeds..i.e.2ab/a+b. might save you precious 30-40 seconds.

Re: least distance [#permalink]
22 Nov 2008, 08:02

Do you think this original question is beyond what GMAT authors may write for a question? It's a great practice question, but I think it's harder than a typical GMAT question. Maybe this would be a 750+ question.
_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: least distance [#permalink]
22 Nov 2008, 08:12

1

This post received KUDOS

yeah...may or may not be...as atletikos proved...it can be solved under 2 mins if you can visualise it the 1st time...although i took close to 3-3.5 mins to decide the approach...

jallenmorris wrote:

Do you think this original question is beyond what GMAT authors may write for a question? It's a great practice question, but I think it's harder than a typical GMAT question. Maybe this would be a 750+ question.

Re: least distance [#permalink]
22 Nov 2008, 08:15

Lol. You could use a calendar to time how long I took!

prasun84 wrote:

yeah...may or may not be...as atletikos proved...it can be solved under 2 mins if you can visualise it the 1st time...although i took close to 3-3.5 mins to decide the approach...

jallenmorris wrote:

Do you think this original question is beyond what GMAT authors may write for a question? It's a great practice question, but I think it's harder than a typical GMAT question. Maybe this would be a 750+ question.

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: least distance [#permalink]
22 Nov 2008, 18:02

ritula wrote:

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

1.4 sqrt(2) 1.7 sqrt(3) 2

A. Agree with A. very good question with good concepts.

I would draw a perpendicular line from the origin to line y = (3/4)x - 3 and then get the length of that perpendicular line, whch is 2.4. Then length of perpendicular line minus 1 (length of redius) is equal to 1.4.

Drawing a picture would be much useful in visualizing the required properties..
_________________

Re: least distance [#permalink]
01 Dec 2008, 10:09

A circle is a series of points that are equal distance from a single point. That single point is the origin. Here, the problem tells the exact location of that point, (0,0). The radius of 1 means that each point that makes up the circle is 1 unit from (0,0).

Does that help?

mbaobsessed wrote:

jallenmorris, prasun84 and others,

Kudos to you all for explaining this.

But I could not even start this. coz I don't understand how all of you arrived at this.

Circle with (0,0) origin and radius 1 ??

Can anyone explain this part in detail pls ?

-mo

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: least distance [#permalink]
20 Apr 2010, 04:25

6

This post received KUDOS

Common guys, it's not as tough as seems to be... just easy, X^2+Y^2=1 is circle with center (0,0) and radius 1, just simply x=0, y= +/-1, and y=0, x=+/-1, and y=3/4-3 is just a line, which forms right triangle(3,4,5) with origin. Assume that Z is perpendicular to hypothenuse, r is radius of circle, which is 1, d is Distance from our circle to line. so Area of this triangle is 1/2*3*4=1/2*5*Z, so Z is equal to 12/5. And, Z=r+d=>12/5=1+d=> d=7/5=1.4

Re: least distance [#permalink]
20 Apr 2010, 08:18

gmatbull wrote:

ritula wrote:

I tried calculating OD from Tri OBD: Using 30:60:60 degrees OD = 2, since OB (side facing 90 deg) = 4 Least distance = 2 – 1 = 1

Using Tri OAD” Again 30:60:90 (x=3/2) OD = 3/2(\sqrt{3}) OD = 2.5 Least distance = 2.5 – 1 = 1.5 Different answers!

It's not a 30:60:90 triangle. The original triangle (your OAB) is a pythagorean triple 3:4:5. You can find OD by getting the height of the original triangle, using AB as a base.

gmatclubot

Re: least distance
[#permalink]
20 Apr 2010, 08:18