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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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11 Oct 2009, 18:19
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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 = 3/4*x - 3?

A. 1.4

B. $$\sqrt{2}$$

C. 1.7

D. $$\sqrt{3}$$

E. 2.0
[Reveal] Spoiler: OA

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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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12 Oct 2009, 00:54
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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.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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12 Oct 2009, 06:45
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Lets do in 1 min.

Use the formula

D = | Am+Bn+C|/ SQRT(A^2 + B^2) where Ax+By+C = 0

put (m,n) =0,0 = center of circle

we get D = 12/5 thus required distance is D-1 = 12/5 -1 = 7/5 = 1.4

we don't require points
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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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12 Oct 2009, 09:28
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Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html

You can continue discussions and see the solutions there.

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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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14 Oct 2009, 05:23
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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.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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14 Dec 2009, 20:55
gurpreetsingh wrote:
Lets do in 1 min.

Use the formula

D = | Am+Bn+C|/ SQRT(A^2 + B^2) where Ax+By+C = 0

put (m,n) =0,0 = center of circle

we get D = 12/5 thus required distance is D-1 = 12/5 -1 = 7/5 = 1.4

we don't require points

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

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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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15 Dec 2009, 06:58
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ISBtarget wrote:
I am too weak in this section, could you please explain...what is this formula and why did you do D-1?

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

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.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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15 Dec 2009, 13:50
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?

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

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

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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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17 Dec 2009, 13:55
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

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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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17 Dec 2009, 14:11
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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: $$\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}$$ --> $$\frac{height}{3}=\frac{4}{5}$$ --> $$height=2.4$$.

$$Distance=height-radius=2.4-1=1.4$$

Hope it's helps.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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31 Dec 2010, 08:45
Have questions based on this concept (i.e. Circle on a plane) ever been tested by the GMAT?
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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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31 Dec 2010, 09:00
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rahul321 wrote:
Have questions based on this concept (i.e. Circle on a plane) ever been tested by the GMAT?

Yes. Check the following GMAT prep questions:
point-on-a-circle-106016.html
does-line-k-touch-circle-or-not-101471.html
plane-geometry-semicircle-from-gmatprep-85154.html
does-the-point-p-touch-the-circle-101485.html
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Re: Tough and tricky 7: distance between the circle an line [#permalink]

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31 Dec 2010, 10:17
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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 22070 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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17781 [17], given: 235 Math Expert Joined: 02 Sep 2009 Posts: 42249 Kudos [?]: 132626 [0], given: 12326 Re: Tough and tricky 7: distance between the circle an line [#permalink] ### Show Tags 31 Dec 2010, 10:23 Expert's post 2 This post was BOOKMARKED 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 _________________ Kudos [?]: 132626 [0], given: 12326 Manager Joined: 23 Oct 2009 Posts: 72 Kudos [?]: 35 [0], given: 14 Re: Tough and tricky 7: distance between the circle an line [#permalink] ### Show Tags 01 Jan 2011, 03:08 Bunuel wrote: rahul321 wrote: Have questions based on this concept (i.e. Circle on a plane) ever been tested by the GMAT? Yes. Check the following GMAT prep questions: point-on-a-circle-106016.html does-line-k-touch-circle-or-not-101471.html plane-geometry-semicircle-from-gmatprep-85154.html does-the-point-p-touch-the-circle-101485.html Thanks. I'm just reading the GMATclub Coordinate Geometry post now...Great stuff!! It's surprising that the OG Quant book doesn't remotely touch upon the 'Circle on a plane' topic even though it is tested! _________________ Consider giving Kudos if you liked my post. Thanks! Kudos [?]: 35 [0], given: 14 Manager Joined: 18 Jun 2010 Posts: 142 Kudos [?]: 38 [1], given: 2 Re: Tough and tricky 7: distance between the circle an line [#permalink] ### Show Tags 22 Oct 2011, 11:35 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. Brilliant explanation. Kudos [?]: 38 [1], given: 2 Intern Joined: 20 Aug 2010 Posts: 7 Kudos [?]: 9 [0], given: 1 Schools: Duke,Darden,Chicago University M13 Q12 [#permalink] ### Show Tags 04 Dec 2011, 19:56 1 This post was BOOKMARKED What is the least possible distance between a point on the circle $$x^2 + y^2 = 1$$ and a point on the line $$y = \frac{3}{4}x - 3$$ ? $$1.4$$ $$\sqrt{2}$$ $$1.7$$ $$\sqrt{3}$$ $$2.0$$ Can anyone Please explain this Question?? Kudos [?]: 9 [0], given: 1 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7736 Kudos [?]: 17781 [1], given: 235 Location: Pune, India Re: M13 Q12 [#permalink] ### Show Tags 05 Dec 2011, 00:45 1 This post received KUDOS Expert's post CracktheGmat2010 wrote: What is the least possible distance between a point on the circle $$x^2 + y^2 = 1$$ and a point on the line $$y = \frac{3}{4}x - 3$$ ? $$1.4$$ $$\sqrt{2}$$ $$1.7$$ $$\sqrt{3}$$ $$2.0$$ Can anyone Please explain this Question?? The question has been discussed before. This is my take on it. 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? (shown by the bold line) Attachment: File.jpg [ 17.76 KiB | Viewed 3186 times ] 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. Can we say it will be the least in case of the bold line which is perpendicular to the given line? Yes, it will be because in all other cases, the lines are longer than the perpendicular and hence (line - 1) will also be longer. Then, let's try to find the length of the bold line, x. Since hypotenuse is 5, (1/2)*3*4 = (1/2)*5*x = Area of triangle made by the co-ordinate axis and the given line x = 2.4 So minimum distance is 2.4 - 1 = 1.4 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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07 Dec 2011, 08:15
VeritasPrepKarishma wrote:
CracktheGmat2010 wrote:
What is the least possible distance between a point on the circle $$x^2 + y^2 = 1$$ and a point on the line $$y = \frac{3}{4}x - 3$$ ?

$$1.4$$
$$\sqrt{2}$$
$$1.7$$
$$\sqrt{3}$$
$$2.0$$

Can anyone Please explain this Question??

The question has been discussed before. This is my take on it.

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? (shown by the bold line)

Attachment:
File.jpg

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. Can we say it will be the least in case of the bold line which is perpendicular to the given line? Yes, it will be because in all other cases, the lines are longer than the perpendicular and hence (line - 1) will also be longer.

Then, let's try to find the length of the bold line, x.
Since hypotenuse is 5,
(1/2)*3*4 = (1/2)*5*x = Area of triangle made by the co-ordinate axis and the given line
x = 2.4
So minimum distance is 2.4 - 1 = 1.4

i have one serious doubt
if we find the perpendicular distance (d) given by the formula
|ax+by+c|/(a^2+b^2)^1/2

the value comes to be 4 against your value which comes out to be 2.4
tell me where i am going wrong.

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13 Dec 2011, 00:15
what is the difficulty level of this question?

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Re: M13 Q12   [#permalink] 13 Dec 2011, 00:15

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