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DS : Coordinates with Circles - HELP

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by soni_pallavi » Mon Oct 29, 2012 12:55 am
Hi

Could someone plz explain this DS problem???

In the XY plane point r,s lies on the circle with center at the origin.What is the value of (r^2 + s^2)

1)Radius is 2

2)The point (2^1/2 , -2^1/2) lies on the circle

Thanks
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Source: — Data Sufficiency |
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by anuprajan5 » Mon Oct 29, 2012 1:21 am
The answer is D

Since r,s lies on the circle with center at origin, distance between the point and center is the radius. root ((r-0)^2+(s-0)^2) = root(r^2+s^2)

Statement 1 - Radius is 2. We can determine r^2+s^2. Sufficient

Statement 2 - The point (2^1/2 , -2^1/2) lies on the circle. The distance between this point and the center is equal to the distance between R,S and the center. Sufficient.
Regards
Anup

The only lines that matter - are the ones you make!

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by gaurvipul » Mon Oct 29, 2012 5:45 am
This is a rephrasing question.

You can simply rephrase it to, What is radius of the circle having center at origin?

1)Radius is 2
if we draw perpendicular from point (r,s) x-intercept would be r and y-intercept would be s.

r^2+s^2 would be square of the hypotenuse which happens to be the radius. i.e. 2^2 = 4.

SUFFICIENT

2)The point (2^1/2 , -2^1/2) lies on the circle
Let us consider this point as (u,v). Again, drawing perpendiculars, we get the x & y intercept both of which have value = 2^(1/2)

Again, hypotenuse is the radius which will remain same whether it is this point or (r,s).

so,r^2+s^2 = (2^(1/2))^2+(2^(1/2))^2=4

SUFFICIENT
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