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What is the least possible distance between a point on the

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Veritas Prep GMAT Instructor
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Re: M13 Q12 [#permalink]

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13 Dec 2011, 02:28
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sandeeepsharma wrote:
what is the difficulty level of this question?

700+, actually closer to 750.
It takes a degree of imagination to figure out the solution quickly.
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Re: What is the least possible distance between a point on the [#permalink]

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

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10 Jul 2014, 07:19
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For the people still troubled with this questions -

First of all this questions tests relatively advanced skills in mathematics and hence IMO can not be a part of the real GMAT.

However, to find the answer of this Q we need to follow below mentioned steps:

1) find the distance of the line from origin : this distance should be the shortest possible distance
2) as all points on circle are equi distance from the origin, we need to find the shortest distance of line from origin and subtract radius from it to get our answer
3) to get shortest distance we need to actually find length of perpendicular line which starts from origin and ends at our given line

3-a) One of the method to solve for (3) is using the equation - |ax1 + by1 + c| /sqrt (a^2+b^2) (this formula you need to remember) - read posts by bunuel or gurpreet for more details.
3-b) Another method is to find the equation of perpendicular line and then find an intersection point of this perpendicular line with our given line. Now find the distance between this point to origin (PHEW) - I surely can't do all this in less 2 minutes and be accurate to the second decimal point [ remember our options are 1.4 and sqrt (2) = 1.41 ]

Finally, for the purpose of GMAT only, i would advise you should not be worried if you can't remember this formula or find this question too difficult.
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Re: What is the least possible distance between a point on the [#permalink]

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27 Jul 2015, 12:15
Hi Bunuel
Engr2012
Can you please tell why did we took a=-3/4 and b=1
From what we got the values of a and b

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

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27 Jul 2015, 12:59
Shree9975 wrote:
Hi Bunuel
Engr2012
Can you please tell why did we took a=-3/4 and b=1
From what we got the values of a and b

Thanks

If you note carefully, that the formula written by Bunuel above for the shortest distance =

Line: $$ay+bx+c=0$$, point $$(x1,y1)$$

$$d=\frac{|ay1+bx1+c|}{\sqrt{a^2+b^2}}$$

So what it means is that the shortest distance of this line (ax+by+c=0) from (x1,y1) which is the origin (0,0) in our case is

$$d=\frac{|c|}{\sqrt{a^2+b^2}}$$. This is obtained by putting x1=0, y1=0,

For a and b, you need to bring the given equation of line :$$y = \frac{3}{4}*x-3$$ in the form $$ay+bx+c=0$$ to get : $$-4y+3x-12=0$$. Comparing these 2 equations we get,

a =-4, b =3 and c = -12

Substitute the above values of a,b,c,x1,y1 in the formula for d above to get:

$$d=\frac{|-12|}{\sqrt{(-3)^2+(4)^2}}=\frac{12}{5}=2.4$$ (Bunuel had divided this equation all across by 4. But you can also stick to this method).

And then the final required distance = 2.4 - radius of the circle = 2.4-1 = 1.4.

An alternate solution is below:

Refer to the attached figure for details of the variables used. You can see that the shortest distance to the line will be the perpendicular drawn from the center of the circle.

Also, in triangle POQ, right angled at O and OR is perpendicular to PQ.

In right triangle POQ , $$PQ^2=OP^2+OQ^2$$ ---> $$PQ^2 = 3^2+4^2$$ ---> $$PQ=5$$

Area of triangle POQ = $$0.5*OP*OQ = 0.5*OR * PQ$$ ---> $$OR = \frac{OP*OQ}{PQ}$$ ---> $$OR = \frac{3*4}{5}$$ ---> $$OR = 2.4$$

Finally,$$OR = OS + SR$$, OS = radius of the circle = 1 ---> $$SR = OR - OS = 2.4 -1 = 1.4$$. A is the answer.
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Re: What is the least possible distance between a point on the [#permalink]

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20 Apr 2017, 11:55
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

your solution will totally break down if the line cut the circle. In that case, the distance between 1 point ON the circle and the line is 0
Re: What is the least possible distance between a point on the   [#permalink] 20 Apr 2017, 11:55

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