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...
xy plane
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
-
mike22629
- Master | Next Rank: 500 Posts
- Posts: 322
- Joined: Fri Mar 27, 2009 3:56 pm
- Thanked: 24 times
- GMAT Score:710
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
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
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...
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...
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...
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...
