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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

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: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) --> \(\frac{height}{3}=\frac{4}{5}\) --> \(height=2.4\).

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.

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 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. \(2.0\)

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: \(2.4-1(radius)=1.4\)

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

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.

Re: What is the least possible distance between a point on the [#permalink]
07 Jan 2013, 11:36

1

This post received KUDOS

1

This post was BOOKMARKED

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

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.

Re: What is the least possible distance between a point on the [#permalink]
16 May 2013, 23:52

2

This post received KUDOS

Attachment:

untitled.JPG [ 6.09 KiB | Viewed 3074 times ]

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 _________________

"Kudos" will help me a lot!!!!!!Please donate some!!!

Completed Official Quant Review OG - Quant

In Progress Official Verbal Review OG 13th ed MGMAT IR AWA Structure

Yet to do 100 700+ SC questions MR Verbal MR Quant

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

Re: What is the least possible distance between a point on the [#permalink]
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). _________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

Re: What is the least possible distance between a point on the [#permalink]
04 May 2014, 01:39

Expert's post

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).

The Importance of Financial Regulation : Before immersing in the technical details of valuing stocks, bonds, derivatives and companies, I always told my students that the financial system is...

One question I get a lot from prospective students is what to do in the summer before the MBA program. Like a lot of folks from non traditional backgrounds...