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...

Hi augusto,

Thanks for thr reply. No, there's no typo in second sentence!. It is as it says.

Upz,

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

Thanks,
augusto

It stands for square root

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

Typo in B.

s^2 + r^2 = 4

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...

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...