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# PS -GMATprep -circle

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VP
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22 Jun 2008, 08:59
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circle-coordinate-diag.doc [74.5 KiB]

Last edited by goalsnr on 22 Jun 2008, 11:36, edited 1 time in total.
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22 Jun 2008, 09:01
i cant open the attchment..can you repost..
VP
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22 Jun 2008, 11:37
fresinha12 wrote:
i cant open the attchment..can you repost..

I deleted the existing file and uploaded once gain.Pls let me know if you still unable to download teh file.
Intern
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22 Jun 2008, 11:48
The correct answer is 1. apply the 30 - 60 -90 rule and radius and diagonal is 2.
VP
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22 Jun 2008, 13:25
abhitheCEO wrote:
The correct answer is 1. apply the 30 - 60 -90 rule and radius and diagonal is 2.

The distance between O and P gives the radius of the circle.

distance = sqrt ( (x1-x2)^2 + (y1-y2)^2)
radius = sqrt ( 3+1) = 2

How did you manage to get the radius and diagonal of the same length
VP
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22 Jun 2008, 13:43
Ok i figured it out .

I 'll wait for sometime before I post my explanation incase others want to try.
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22 Jun 2008, 19:36
goalsnr wrote:
Ok i figured it out .

I 'll wait for sometime before I post my explanation incase others want to try.

I think as a general rule, when the lines are of the same length (radius in this case) and they are perpendicular to each other ...... they are mirror images of each other and their co-ordinates are merely interchanged hence (-sqrt3, 1) would become (1,sqrt3).

If you have a better explanation i would love to know it !!
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23 Jun 2008, 01:48
goalsnr wrote:

From the document, we all know the answer is 1 .

Here's how - OP, we can calculate to be 2. So OPQ is a right triangle with sides, 2,2,x (x being hypotenuse, PQ).

PQ = sqrt(4+4) = 2 * sqrt(2).

So s = PQ - horizontal distance of Q from y-axis

ie,
s = 2 * sqrt(2) - sqrt(3) (since x-value of P is sqrt(3).

sqrt(2) ~ 1.4; sqrt(3) ~ 1.7

Hence, s = 2.8 - 1.7 ~= 1.

Sorry if the symbols are confusing!
SVP
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23 Jun 2008, 03:52
Here is my explanation.

Since OP and OQ are perpendicular to each other, the multiplication of their slopes should be -1.

Thus (1-0)/(-v3-0) * (t-0)/(s-0) = -1
That means, t = v3s

or 3 + 1 = s^2 + t^2

Replace value for t and this will give s = 1.
VP
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23 Jun 2008, 17:10
saurabhkowley18 wrote:
goalsnr wrote:
Ok i figured it out .

I 'll wait for sometime before I post my explanation incase others want to try.

I think as a general rule, when the lines are of the same length (radius in this case) and they are perpendicular to each other ...... they are mirror images of each other and their co-ordinates are merely interchanged hence (-sqrt3, 1) would become (1,sqrt3).

If you have a better explanation i would love to know it !!

:D thats the trap I fell for. There is a classic parabola question using the mirror image and I thought this question was on the similar lines.
VP
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23 Jun 2008, 17:19
scthakur wrote:
Here is my explanation.

Since OP and OQ are perpendicular to each other, the multiplication of their slopes should be -1.

Thus (1-0)/(-v3-0) * (t-0)/(s-0) = -1
That means, t = v3s

or 3 + 1 = s^2 + t^2

Replace value for t and this will give s = 1.

Nice ! I solved using triangle info.See explanation below
VP
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23 Jun 2008, 17:23
Draw PM and QN perpendicular to the diagonal.

Triangle PMO i: PM = 1, PO= radius =2
its a 30:60:90 - triangle.
angle POM = 30.

Using this info angle QON is 60
30:60:90
1:sqrt 3 :2

ON =s =1
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23 Jun 2008, 18:42
so i just learned this trick the other day...if you have two lines intersecting to form a righ-angle triangle..you can simply use the formula, slope Line1*slope Line2=-1..

basing that slope of line 1 is -sqrt(3)/1*1/sqrt(3)..means s=1..
VP
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23 Jun 2008, 20:57
if 2 lines are perpendicular to each other the product of their slopes is -1
->m1= -1/m2
if 2 lines are parallel to each other the product of their slopes are equal
->m1=m2
Re: PS -GMATprep -circle   [#permalink] 23 Jun 2008, 20:57
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