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What is the least possible distance between a point on the

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Re: least distance [#permalink]  20 Apr 2010, 08:58
I think a lot of people are making this question way harder than it has to be.

Start by graphing the circle at (0,-1), (1,0), (0,1), (,-1,0)

Then graph the line at (0,-3) and (4,0)

What is the area? Simple...6

Now figure out hypotenuse of the 3, 4, 5 triangle. Are should still be 6 because it is the same triangle. Use 5 as your base and figure out h= 2.4. Now minus the 1 from the circle and it equals 1.4

20 sec problem easy.
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Re: least distance [#permalink]  20 Apr 2010, 09:04
mmphf wrote:
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

Again 30:60:90 (x=3/2)
OD = 3/2(\sqrt{3})
OD = 2.5
Least distance = 2.5 – 1 = 1.5

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.

Thanks, i got it now.
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Re: least distance [#permalink]  20 Apr 2010, 12:26
fully agree with ksharma.

Just plot the points on x and y-axis for the two equations (1,1,-1,-1...4,-3 ) and you'll easily find the solution from the height of the 3,4,5 triangle.

easy under 1min question.
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Re: least distance [#permalink]  20 Apr 2010, 13:53
Only 10% of Gmat takers will be able to get to it in less than 1.5 mins. I think in real GMAT, educated guess is the best way to approach this.
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Re: least distance [#permalink]  20 Apr 2010, 14:58
Least distance between the circle and the given line will be the perpendicular distance between the given line and the tangent at a point on the circle where it is parallel to the given line.

Thus the required distance = Distance of given line from origin - radius

= 2.4-1 = 1.4

Reason : Line drawn form the center to tangent will be perpendicular and now if we extend it to the given line it will be perpendicular to the given line as well.
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Re: least distance [#permalink]  20 Apr 2010, 20:11
basik21 wrote:
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

Good one , really simple method , thanks
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Re: least distance [#permalink]  21 Apr 2010, 04:48
Area of triangle = 1/2 base * height
=> 1/2 * 3 * 4 = 1/2 * 5 * perpendicular on hypotenuse
=> perpendicular = 2.4

=> distance from circle = 1.4 (i.e. 2.4 - radius)
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Re: least distance [#permalink]  15 Jun 2010, 19:30
Thanks to all the guys who provided the solution.
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Re: least distance [#permalink]  06 Sep 2010, 17:39
Prasun, how r u calculatin this?

the length of the perpendicular from the centre(0,0)= 3/ sq root?
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Re: least distance [#permalink]  06 Sep 2010, 23:13
absolutely amazing explaination
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Re: least distance [#permalink]  22 Apr 2011, 03:34
4
KUDOS

x intercept (OA) = 4

y intercept (OB) = 3

5 - Hypotenuse

1/2 * OD * 5 = 1/2 * 12

=> OD = 12/5

So distance = 12/5 - 1 = 7/5

= 1.4

Attachments

Circle.png [ 8.06 KiB | Viewed 963 times ]

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Re: least distance [#permalink]  22 Apr 2011, 04:03
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 $$y = \frac{3}{4}x - 3$$ ?

(A) $$1.4$$
(B) $$\sqrt{2}$$
(C) $$1.7$$
(D) $$\sqrt{3}$$
(E) $$2.0$$

[Reveal] Spoiler: OA
A

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Re: least distance [#permalink]  22 Apr 2011, 13:05
How do we know the perpendicular is passing through origin?
I could not figure our that. I missed something.

Anyway I used derivative(have some blurred memories of this:) to minimize distance.

Say point (a,b) on circle which is nearest to the line
Now distance between a line and a point is
d = |b-ma-c|/sqrt(m^2+1) where m is the slope of line and c is intercept

= |b-(3/4)a +3|/ sqrt( (3/4)^2 +1)

(a,b) on circle, hence a^2 + b^2 = 1 => b = sqrt(1-a^2)

Use this in distance eqn and minimize the numerator using derivative.
I wont go into that. Just do df/da = 0 and remember a should be + and b should be - as this point is in 4th quad.

So I got a = 3/5 and b = -4/5, this is the point of tangent to the circle parallel to given line

d = 1.4
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Re: least distance [#permalink]  22 Apr 2011, 13:54
Expert's post
hellishbrain wrote:
How do we know the perpendicular is passing through origin?
I could not figure our that. I missed something.

The point is not whether the perpendicular on the line will pass through the center. The point is that the perpendicular from the center will give the minimum distance. Look at the diagram below and forget the circle for the time being. What is the minimum distance from the center (0,0) to the line? It will the perpendicular from the center to the line, right? Now think, what will be the shortest distance from the circle to the line? It will be 1 unit less than the distance from the center to the line, isn't it?

Attachment:

Ques2.jpg [ 17.76 KiB | Viewed 778 times ]

Since hypotenuse is 5,
(1/2)*3*4 = (1/2)*5*x
x = 2.4
So minimum distance is 2.4 - 1 = 1.4
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 20 Jan 2011 Posts: 88 Followers: 3 Kudos [?]: 36 [0], given: 23 Re: least distance [#permalink] 22 Apr 2011, 14:13 Thanks. So take away is for any circle(centered anywhere) and any line, the perpendicular from the centre of circle to line would give the minimum distance. Right? _________________ Conquer the Hell and make it Haven. Brain is your hell and Success is your haven! "Kudos" is significant part of GMAT prep. If you like it, you just click it Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5735 Location: Pune, India Followers: 1444 Kudos [?]: 7572 [0], given: 186 Re: least distance [#permalink] 22 Apr 2011, 14:20 Expert's post hellishbrain wrote: Thanks. So take away is for any circle(centered anywhere) and any line, the perpendicular from the centre of circle to line would give the minimum distance. Right? Certainly! Just forget the circle itself. Think about the point at the center of the circle. If I had to find the minimum distance from that point to a line, it would be the perpendicular distance. So minimum distance of any point on the circle to the line would be that distance minus the radius. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: least distance [#permalink]  22 Apr 2011, 19:05
Thanks for the explanation guys and various methods to solve it..
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Re: least distance [#permalink]  22 Apr 2011, 19:27
Alternately, being a right angle triangle, with base can be taken as either of sides and height will be the perpendicular from vertex left aside.
Area = 0.5 * base * height = 0.5 * 3 * 4 = 6 (here we compute area)
Area = 0.5 * base * height = 0.5 * 5 * h, thus h = 2.4. However, h should be reduced by 1 as it's radius of circle. Hence, distance between circle and line is 1.4.
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Re: least distance [#permalink]  23 Apr 2011, 12:10
atletikos wrote:
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

great work!! it goes to show that there's usually a quick way hiding somewhere in a triangles question
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Re: least distance [#permalink]  23 Apr 2011, 13:04
we can see that this line forms a right angled triangle with origin.

let perpendicular distance from center of the circle/origin to line be h.

by comparing area of the triangle we have

1/2 * 4 * 3 = 1/2 * h * 5

h = 12/5 =2.4

minimum distance from the circle to the line = 2.4 -1 = 1.4
Re: least distance   [#permalink] 23 Apr 2011, 13:04

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