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: Tough and tricky 7: distance between the circle an line [#permalink]
11 Oct 2009, 23:54

Bunuel 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

Good problem again. Takes into account a lot of co ordinate geometry fundas.

eqn of circle = x^2 + y^2 = 1, center = (0,0) radius = 1

min dist of line from circle = dist of line from the center - radius

Make the distance of the line from the circle to be 0 and we see that it becomes a tangent to the circle.

Now we know if we draw a line from the center to the point where the tangent touches the circle the line and the tangent are perpendicular to each other.

So we need to find the equation of this line first.

We can take the line back where it was now

Since the lines are perpendicular m1 x m2 = -1

m of line = 3/4

so slope of the new line = -4/3

Since the line passes through the origin (center of circle) its eqn => y=-4/3x

now we need to get the point of intersection of our two lines, which comes out to be (36/25,-48/25)

now get the distance of this point from the origin and subtract the radius from it.

Comes to 1.4 (may have made calculation errors )

So A.

Comes under 2 mins.

Bunuel, great work with the Questions. I suggest you make a single thread and keep updating it. People can subscribe to that and also it will help new guys read all the probs and solutions in one thread.
_________________

Re: Tough and tricky 7: distance between the circle an line [#permalink]
14 Oct 2009, 04:23

I have learnt this in my school time that distance from one point to a line is what stated above.

Actually its same like finding the line which originates from the distant point and then intersect at the required line, and then we find the point of intersection and then using the formula we calculate distance between the two points.

But even if u dont rem the formula what u can do is....

suppose line is ax+by-c=0 now when y=0 , x=c/a and when x=0 , y = c/b

area of triangle formed by these 2 points and center (0,0) is 1/2 * c/a * c/b = c^2/2ab

now this is equal to 1/2 * D1 * D2 , where D1 is distance between points on x and y coordinates of the line which is sqrt [ (c/a)^2 + (b/a) ^2 ] and D2 is the required perpendicular distance on the line.

Equate them and u will get the ans. Its very easy concept and I m not that good in explaining here I think sry for that.

If you still dont get this please letme know I will explain again.
_________________

Re: Tough and tricky 7: distance between the circle an line [#permalink]
15 Dec 2009, 05:58

4

This post received KUDOS

Expert's post

ISBtarget wrote:

I am too weak in this section, could you please explain...what is this formula and why did you do D-1? please help

First 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).

Now we can do this by finding the equation of a line perpendicular to given line y=\frac{3}{4}*x-3 (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points (x1,y1) and (x2,y2): d=\sqrt{(x1-x2)^2+(y1-y2)^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 (x1,y1)

d=\frac{|ay1+bx1+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.

P.S. Also note that when we have x^2+y^2=k, we have circle (as we have x^2 and y^2), it's centered at the origin (as coefficients of x and y are 1) and the radius of that circle r=\sqrt{k}.

You can check the link of Coordinate Geometry below for more.
_________________

Re: Tough and tricky 7: distance between the circle an line [#permalink]
15 Dec 2009, 12:50

Bunuel wrote:

ISBtarget wrote:

I am too weak in this section, could you please explain...what is this formula and why did you do D-1? please help

First 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).

Now we can do this by finding the equation of a line perpendicular to given line y=\frac{3}{4}*x-3 (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points (x1,y1) and (x2,y2): d=\sqrt{(x1-x2)^2+(y1-y2)^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 (x1,y1)

d=\frac{|ay1+bx1+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.

P.S. Also note that when we have x^2+y^2=k, we have circle (as we have x^2 and y^2), it's centered at the origin (as coefficients of x and y are 1) and the radius of that circle r=\sqrt{k}.

You can check the link of Coordinate Geometry below for more.

Awesome man, why wouldnt you start a quant training program....excellent

Re: Tough and tricky 7: distance between the circle an line [#permalink]
17 Dec 2009, 12:55

Hi

One more question...sounds silly but can you help

why are you doing D-1 , you are calculating the distance between a point on the circle and the line , 0,0 is a point on the circle, why cant 2.4 be the answer

Re: Tough and tricky 7: distance between the circle an line [#permalink]
17 Dec 2009, 13:11

Expert's post

Minimum 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).

(0,0) is not the point on the circle, it's the center of the circle with radius 1 (circle is centered at the origin).

2.4 is the distance from the line to the origin (the center of the circle), so we should subtract the length of the radius to get the distance from the line to the circle.

ANOTHER SOLUTION:

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.

Re: Tough and tricky 7: distance between the circle an line [#permalink]
31 Dec 2010, 09:17

7

This post received KUDOS

Expert's post

A rather nice question and I think Bunuel has already explained sufficient alternative solutions. I just want to add a method of finding the altitude of a right triangle here that I think is particularly neat (discussed in Veritas Geometry book)

Attachment:

Ques2.jpg [ 8.1 KiB | Viewed 9588 times ]

I want to find x here since (x - 1) will be the minimum distance from the circle to the line (as explained above)

Area of the given right triangle = (1/2)*3*4 (3 is base and 4 is altitude)= (1/2)*5*x (5 is base and x is altitude) So x = 2.4

Finding the area of the original triangle in two different ways and equating it will help you find the altitude.
_________________

Re: Tough and tricky 7: distance between the circle an line [#permalink]
31 Dec 2010, 09:23

Expert's post

VeritasPrepKarishma wrote:

A rather nice question and I think Bunuel has already explained sufficient alternative solutions. I just want to add a method of finding the altitude of a right triangle here that I think is particularly neat (discussed in Veritas Geometry book)

Attachment:

Ques2.jpg

I want to find x here since (x - 1) will be the minimum distance from the circle to the line (as explained above)

Area of the given right triangle = (1/2)*3*4 (3 is base and 4 is altitude)= (1/2)*5*x (5 is base and x is altitude) So x = 2.4

Finding the area of the original triangle in two different ways and equating it will help you find the altitude.

Yes, this approach is quite handy. Question about this concept: triangles-106177.html _________________

Re: Tough and tricky 7: distance between the circle an line [#permalink]
22 Oct 2011, 10:35

1

This post received KUDOS

Bunuel wrote:

ISBtarget wrote:

I am too weak in this section, could you please explain...what is this formula and why did you do D-1? please help

First 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).

Now we can do this by finding the equation of a line perpendicular to given line y=\frac{3}{4}*x-3 (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points (x1,y1) and (x2,y2): d=\sqrt{(x1-x2)^2+(y1-y2)^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 (x1,y1)

d=\frac{|ay1+bx1+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.

P.S. Also note that when we have x^2+y^2=k, we have circle (as we have x^2 and y^2), it's centered at the origin (as coefficients of x and y are 1) and the radius of that circle r=\sqrt{k}.

You can check the link of Coordinate Geometry below for more.

Re: Tough and tricky 7: distance between the circle an line [#permalink]
06 Dec 2012, 11:45

Bunuel wrote:

Minimum 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).

(0,0) is not the point on the circle, it's the center of the circle with radius 1 (circle is centered at the origin).

2.4 is the distance from the line to the origin (the center of the circle), so we should subtract the length of the radius to get the distance from the line to the circle.

ANOTHER SOLUTION:

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.

Re: What is the least possible distance between a point on the [#permalink]
07 Feb 2014, 21:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: What is the least possible distance between a point on the [#permalink]
09 Feb 2014, 06:50

Took quite some time(6+ mins) to solve . But I took the same approach as explained by VeritasPrepKarishma. I just hope these ideas click on time. Nice question.
_________________

Click on Kudos if you liked the post!

Practice makes Perfect.

gmatclubot

Re: What is the least possible distance between a point on the
[#permalink]
09 Feb 2014, 06:50