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What is the least possible distance between a point on the

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What is the least possible distance between a point on the [#permalink]

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New post 03 Aug 2008, 23:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0
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Re: Co-ordinate geometry [#permalink]

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New post 03 Aug 2008, 23:27
GMAT TIGER 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 = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0


shortest distance from (0,0) to the line - radius of the circle
=12/5 - 1
= 7/5
= 1.4

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Re: Co-ordinate geometry [#permalink]

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New post 04 Aug 2008, 12:35
x2suresh wrote:
durgesh79 wrote:
GMAT TIGER 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 = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0


shortest distance from (0,0) to the line - radius of the circle
=12/5 - 1
= 7/5
= 1.4


Can you elaborate


The line containing the shortest path b/w a circle and line X is the line that is perpindicular to line X and passes through the center of the circle. To calculate the distance b/w the circle and line X, you take the distance from the center of the circle to line X and subtract out the distance from the center of the circle to the edge of the circle (i.e., the radius).

I did it a long way. Durgesh, I don't know if you have a quicker way. This took me more than 2 minutes.

The line, with x and y axes create a 3-4-5 triangle. The non-leg base of this right triangle extends from (0,0) and intersects the hypotenuse / line at a right angle. This base creates 2 new right triangles with hypotenuses created by the x- and y-axis, with lengths 4 and 3, respectively.

Let a = the length of the upper half of the hypotenuse of length 5 created by the intersection of the line and the non-leg base
Let b = the length of the non-leg base

a^2 + b^2 = 4^2 = 16
(5-a)^2 + b^2 = 3^2 = 9

25 - 10a + a^2 + b^2 = 9
25 - 10a + 16 = 9
10a = 32
a = 16/5

(16/5)^2 + b^2 = 16
b^2 = 16 - 16^2/25 = 16*(1-16/25) = 16*9/25
b = sqrt(16*9/25) = 4*3/5 = 12/5

Distance = 12/5 - 1 = 7/5

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Re: Co-ordinate geometry [#permalink]

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New post 04 Aug 2008, 13:38
Ok I got it now.

shortest distance is perdicular line that passes through circle mid point (0,0) and touches the line

We have to find the point where perdicular and given line intersects.
y = 3/4*x - 3 (given line) --(1)
y =-4/3x +c ( perpedicular line ) --> this line passes through (0,0)
y= -4/3x -->(2)
combine 1 and 2 to get the intersect point
-4/3x = 3/4x-3
--> x= (3 * 12)/25
y = (3 * 12)*(4/3)/25
= (4 * 12)/25
so now distince between (0,0) and ( (3 * 12)/25 , (4 * 12)/25)) ---->(3:4:5 triagnle)
clearly = 5 * 12/ 25 = 12/5

Distance = 12/5 - 1 = 7/5
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Re: Co-ordinate geometry [#permalink]

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New post 04 Aug 2008, 14:01
Another way to look at this which is more intuitive to me:
1. Circle has the same radius in all direction/angle
2. The center of the circle is on the origin
3. The shortest distance between a point and a line is perpendicular to the line
4. Find the shortest distance between the origin and the line (can use the formula or can be deduced as already posted)
5. Subtract the length of the radius
Answer: 7/5

Good question, I liked it.
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Re: Co-ordinate geometry [#permalink]

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New post 04 Aug 2008, 22:07
GMAT TIGER 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 = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0


thanks all. looking for more elegant method. :roll:

oa = 1.4
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Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


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Re: Co-ordinate geometry   [#permalink] 04 Aug 2008, 22:07
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