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Source: — Data Sufficiency |
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by augusto » Sun Jul 27, 2008 12:04 am
Hi CITI,

To understand this problem I imagine that the point (r,s) is a vertice in a triangle. So that triangle has one side of lenght *r* another side of length *s* and the lenght of the hypotenuse equal to the radius of the circle.

So if you can imagine that, then r^2 + s^2 is one of the "sides" of the pythagoras theorem. So this is asking the value of the square of the radius.

So from (1) you know that the answer is 2^2 which is enough.
From (2) you can get the value of the radius, because you know that the circle is centered in the origin.

*note*: Just in case I'm assuming that the 'v's in (v2,-v2) are typos, and that the point is (2,-2)... which funny enough gives a different result both answers...
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by CITI29 » Sun Jul 27, 2008 6:28 am
Hi augusto,

Thanks for thr reply. No, there's no typo in second sentence!. It is as it says.
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by augusto » Sun Jul 27, 2008 9:46 am
Upz,

Well, please tell me if you know what those 'v's stand for.

Thanks,
augusto
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by shkusira » Thu Apr 16, 2009 9:58 am
It stands for square root
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by mike22629 » Thu Apr 16, 2009 12:02 pm
What this question comes down to is whether you can determine any point on the circle. This is because any two points on a circle, when squared and added together will all equal the same thing.

A.
Knowing that the radius is 2 and the origin is the center lets you know that (2,0) is on circle.

Hence r^2 + s^2 = 4
Suff.

B.
Knowing that (sq(2), -sq(2)) is obviously a point on the circle.

Hence r^2 + s^4 = 4
Suff.

IMO D
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by mike22629 » Thu Apr 16, 2009 12:03 pm
Typo in B.

s^2 + r^2 = 4
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by pavan.mpv » Tue Feb 21, 2012 11:06 pm
Just remember the basic formula of the coordinate geometry.

Distance between two points (x1,y1) and (x2,y2) is= sqroot ((x2-x1)2 +(y2-y1)2)

Given: x1,y1 = r,s; x2,y2= 0,0 (since origin)

Now
St 1: radius is 2, i.e. sqroot ((r^2-0)+(s^2-0))=2 -> r2+s2=4 -- suff
St 2: (Sqroot(2), -sqroot(2)) --> sqroot (sqroot(2)^2-0 + (-sqroot(2)^2) = sqroot(4)= 2 distance hence we can again find r^2 +s^2 -- suff

Please excuse me of using confusing terminology... :)
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by pavan.mpv » Tue Feb 21, 2012 11:06 pm
Just remember the basic formula of the coordinate geometry.

Distance between two points (x1,y1) and (x2,y2) is= sqroot ((x2-x1)2 +(y2-y1)2)

Given: x1,y1 = r,s; x2,y2= 0,0 (since origin)

Now
St 1: radius is 2, i.e. sqroot ((r^2-0)+(s^2-0))=2 -> r2+s2=4 -- suff
St 2: (Sqroot(2), -sqroot(2)) --> sqroot (sqroot(2)^2-0 + (-sqroot(2)^2) = sqroot(4)= 2 distance hence we can again find r^2 +s^2 -- suff

Please excuse me of using confusing terminology... :)
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