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What is the least possible distance between a point on the [#permalink] New post 12 Apr 2010, 00:26
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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
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 12 Apr 2010, 04:32
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abhi758 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

Could you please explain the working for arriving at the radius of the circle = 1? Look forward to the replies.


In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
(x-a)^2+(y-b)^2=r^2

If the circle is centered at the origin (0, 0), then the equation simplifies to:
x^2+y^2=r^2

So, the circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1.

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line y = \frac{{3}}{{4}}x-3.

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> leg_1=4.
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> leg_2=3.

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \frac{height}{leg_1}=\frac{leg_2}{hypotenuse} --> \frac{height}{3}=\frac{4}{5} --> height=2.4.

Distance=height-radius=2.4-1=1.4

Answer: A.
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 12 Apr 2010, 05:10
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I have a different approach to this. I first checked the options and there is 15-20% differences in the values of all optoins.

From x2 + y2 = 1, we have center at (0,0). Now, for the shortest distance b/w two points is the perpendicular distance.

The slope of line | to line y=3x/4 -3 will be (-4/3) and as this | is in line with the center of circle, so its line eq => y = -4x/3

Now, just equating the two line equations to solve for x and y, we get:
x=36/25 and y=48/25. I assumed the closest values from these ratios as:
x=3/2 and y=2

This is for quick calculation. So, the distance b/w (0,0) and (3/2,2) is nearly \sqrt{(25/4)} is 2.5. We assumed the ratio above (x=36/25 = 3/2 and y=48/25 = 2) a bit more. So, this number should be a bit less. So, I took it as 2.4.

You have noticed that the radius is 1. Thus, the distance b/w a point at circle and the first line will be 2.5 -1 = 1.4, which is A.

I hope this helps.
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 12 Apr 2010, 05:36
Thanks ykaiim and Bunnel for the explanations..

Bunnel, you provide some of the best explanations for the Quant problems..:) Thanks again!
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 22 Sep 2010, 00:37
Bunuel wrote:
abhi758 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

Could you please explain the working for arriving at the radius of the circle = 1? Look forward to the replies.


In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
(x-a)^2+(y-b)^2=r^2

If the circle is centered at the origin (0, 0), then the equation simplifies to:
x^2+y^2=r^2

So, the circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1.

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line y = \frac{{3}}{{4}}x-3.

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> leg_1=4.
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> leg_2=3.

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \frac{height}{leg_1}=\frac{leg_2}{hypotenuse} --> \frac{height}{3}=\frac{4}{5} --> height=2.4.

Distance=height-radius=2.4-1=1.4

Answer: A.

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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 22 Sep 2010, 07:19
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Equation for line can be written as : 4y-3x+12 = 0
Some basic equations :

Distance between any point ( x1, y1 ) and line ax+by+c=0 is

ax1+by1+c/ \sqrt{a2+b2}

Here (x1,y1 ) is (0,0) ( center of circle )
ANd from equation of line , a=3 b=4 and c =12
Hence distance between center of circle and line will be :

= 4*0 + 3*0 + 12/ \sqrt{4*4 + 3* 3}
= 12/\sqrt{5}
=2.4

Distance between center of circle and line is 2.4 .

SInce radius is 1 , least distance between line and circle will be 2.4-1 = 1.4

Hope this helps
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 22 Sep 2010, 07:41
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msbinu wrote:
Equation for line can be written as : 4y-3x+12 = 0
Some basic equations :

Distance between any point ( x1, y1 ) and line ax+by+c=0 is

ax1+by1+c/ \sqrt{a2+b2}

Here (x1,y1 ) is (0,0) ( center of circle )
ANd from equation of line , a=3 b=4 and c =12
Hence distance between center of circle and line will be :

= 4*0 + 3*0 + 12/ \sqrt{4*4 + 3* 3}
= 12/\sqrt{5}
=2.4

Distance between center of circle and line is 2.4 .

SInce radius is 1 , least distance between line and circle will be 2.4-1 = 1.4

Hope this helps


Yes, this question can be solved this way too, though I've never seen any GMAT question requiring the formula used in it. Also there is a little mistake in formula you posted: nominator should be in absolute value.

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

y = \frac{{3}}{{4}}x-3 --> y-\frac{{3}}{{4}}x+3=0.

We know the formula to calculate the distance between two points (x_1,y_1) and (x_2,y_2): d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT:
Line: ay+bx+c=0, point (x_1,y_1)

d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}

DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is (0,0) -->

d=\frac{|c|}{\sqrt{a^2+b^2}}

So in our case it would be: d=\frac{|3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4

So the shortest distance would be: 2.4-1(radius)=1.4

Answer: A.
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 12 Jul 2011, 05:13
can any one explain me THe values of C,a,b---where we get these values in bunnel solution !c!/suare root a^2+b^2(fourmula distance between line and point).
I am confused the value of c,a,b.
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 14 Jul 2011, 19:45
A for me. I solved the equation of the perpendicular line and the circle. Got two points on the circle which were closet to the line, and then used the formula for perpendicular distance from a point to the line to calculate the distance.
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Re: What is the least possible distance between a point on the [#permalink] New post 07 Jan 2013, 11:36
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area of triangle with height as x axis intercept of the line and base as y axis intercept of the line = 1/2*4*3 = 6

Are of the same triangle with base as the segment of the line between x and y intercepts and height as line segment between origin and perpendicular bisector to the line mentioned in the ques = 1/2*h*5

hence h = 12/5 as are will be the same.
h = 2.4
shortest distance = 2.4 - 1 = 1.4
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Re: What is the least possible distance between a point on the [#permalink] New post 22 Feb 2013, 22:33
\frac{height}{leg1}=\frac{leg2}{hypotenuse}

Could someone please explain this
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 08 May 2013, 22:44
rojans wrote:
can any one explain me THe values of C,a,b---where we get these values in bunnel solution !c!/suare root a^2+b^2(fourmula distance between line and point).
I am confused the value of c,a,b.
Thanks


The original line equation is y = 3/4x -3.
if you multiply equation by 4 to both side (to get rid of the fraction) it is still the same equation. In fact it is still same equation if you multiply both sides by any other number (except zero).

equation becomes 4y = 3x-12.
bring all parts to one side,
4y -3x +12 = 0
a=4, b = -3 and c = 12.

EDIT:
You can also use the original equation and write it as y -3/4x + 3= 0. So, a =1, b = -3/4 and c = 3. Bunnel used these in his post.

Last edited by AbuRashid on 08 May 2013, 23:12, edited 1 time in total.
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Re: GMATClub Hardest Questions - Co-Geometry (Q1) [#permalink] New post 08 May 2013, 22:47
msbinu wrote:
Equation for line can be written as : 4y-3x+12 = 0
Some basic equations :

Distance between any point ( x1, y1 ) and line ax+by+c=0 is

ax1+by1+c/ \sqrt{a2+b2}

Here (x1,y1 ) is (0,0) ( center of circle )
ANd from equation of line , a=3 b=4 and c =12
Hence distance between center of circle and line will be :

= 4*0 + 3*0 + 12/ \sqrt{4*4 + 3* 3}
= 12/\sqrt{5}
=2.4

Distance between center of circle and line is 2.4 .

SInce radius is 1 , least distance between line and circle will be 2.4-1 = 1.4

Hope this helps


Many thanks. Something new and very much useful.
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Re: What is the least possible distance between a point on the [#permalink] New post 16 May 2013, 23:17
Another approach to this question: it took me 20 secs to get to the answer.

The line passes through pts (4,-3) forming a right angle triangle with hypotenuse 5 .

The area of the right triangle is (1/2*b*h =6).

We need to calculate the ht from pt 0,0 to the line slope line joining pts (4,-3), let us assume it to be ht

Area =1/2*b*ht , From the previous calculation the area of triangle is 6 and base 5, we can calculate ht which is 12/5

the least possible distance is ht - radius of the circle i.e is 1

Therefore : (12/5) -1 =1.4 , hence A
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Re: What is the least possible distance between a point on the [#permalink] New post 16 May 2013, 23:52
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You can see that AE is the perpendicular distance from the origin .
AF is the radius of the circle.

Theory:- The circle is located at the origin. So the least distance from the origin to the line will be perpendicular distance. Now the least distance from the line to any point on the circle will be this perpendicular distance minus the radius of the circle.


The perpendicular distance from the line to the origin is |C| /(\sqrt{a^2 + b^2}).(if the line is ax + by + c =0)
This value will be 2.4.
So the answer is 2.4 - 1 = 1.4
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Re: What is the least possible distance between a point on the [#permalink] New post 28 Dec 2013, 06:59
abhi758 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


Very nice question indeed, congrats to GMAT club once again

Now let's solve this

1. Draw the line.
2. Now we need the distance from the line to the edge of the circle
3. Note that we have r = 1, and we have a pythagorean triple namely 3,4,5
4. So area of triangle will be 6 which should be equal to b*h / 2 ==> 5x/2 = 6 ===> x = 12/5
5. Now the distance we are looking for is 12/5 - 1 = 7/5 = 1.4

A is your buddy today

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Re: What is the least possible distance between a point on the [#permalink] New post 03 May 2014, 23:47
Hi guys

i am following Manhattan books for prep... no where did i find a line of the circle.. so i got stuck at point 1 only that circle radius is 1 and points 0,0...
where can i get formulas like mentioned by you all, esp bunnel:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
(x-a)^2+(y-b)^2=r^2

If the circle is centered at the origin (0, 0), then the equation simplifies to:
x^2+y^2=r^2

So, the circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1.

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).
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Re: What is the least possible distance between a point on the [#permalink] New post 04 May 2014, 01:39
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nandinigaur wrote:
Hi guys

i am following Manhattan books for prep... no where did i find a line of the circle.. so i got stuck at point 1 only that circle radius is 1 and points 0,0...
where can i get formulas like mentioned by you all, esp bunnel:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
(x-a)^2+(y-b)^2=r^2

If the circle is centered at the origin (0, 0), then the equation simplifies to:
x^2+y^2=r^2

So, the circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1.

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).


I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Also, check my signature below for theory and questions.

Hope this helps.
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Re: What is the least possible distance between a point on the [#permalink] New post 04 May 2014, 03:21
hi Bunnel

I was going through the number properties notes and came across:

If n is a positive integer greater than 1, then there is always a prime number p with n < p < 2n.

Please explain what is meant by this.
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Re: What is the least possible distance between a point on the [#permalink] New post 05 May 2014, 01:45
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nandinigaur wrote:
hi Bunnel

I was going through the number properties notes and came across:

If n is a positive integer greater than 1, then there is always a prime number p with n < p < 2n.

Please explain what is meant by this.


For any integer n greater than 1, there is always at least one prime p such that n < p < 2n (Bertrand's postulate).

It means that you will always find a prime number between n and 2n. For example:

If n=2 then 2 < (3=prime) < 2*2;
If n=3 then 3 < (5=prime) < 2*3;
If n=4 then 4 < (5=prime) < (7=prime) < 2*4;
...

Hope it's clear.
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