circle!

[brick2009]

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

a.) Circle has radius of 2

b.) The point (v2, -v2) Lies on the circle.

Can anyone decipher this que???


oa: D[/spoiler]

[Rahul@gurome]

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

a.) Circle has radius of 2

b.) The point (v2, -v2) Lies on the circle.

Can anyone decipher this que???

oa: D[/spoiler]

(1) This statement implies that the point (2, 0) lies on the circle. So the value of r^2 + s^2 = 2^2 or r^2 + s^2 = 4.
So, (1) is SUFFICIENT.

(2) For this statement can you please check the point again?

[ru2008]

Rahul- Do you mean (2)^2 + (0)^2 when the point is (2,0) or (-2,0)


OR (0)^2 + (2)^2 when the point is (0,2) or (0,-2)

Considering all cases where the point may lie..

[Brian@VeritasPrep]

Hey guys,

This question is in the OG Quant Review book, right? I remember seeing this pretty recently.

As Rahul notes, a circle with a radius of 2 and centered on the origin enables a point like (2, 0) to be on it. In that case, r would be 2 and s would be 0, making r^2 + s^2 = 2^2 + 0^2 = 4.

Now, 2, 0 is an easy point to pick, so you may want to see if the same holds with a different point on the circle. To do that, it's helpful to note that Coordinate Geometry questions on the GMAT allow for us to use right triangles almost always - it's a grid, which means that there are right angles all over the place! If the radius is 2, we can use Pythagorean Theorem to create a right triangle between the origin and the point on the circle, and then use the radius as the hypotenuse (much easier to draw than to explain, so feel free to jot this down!).

If we do that, then a^2 + b^2 will equal the hypotenuse-squared, or 2^2. So we can prove, actually, that the x and y coordinates, which would account for the side lengths of the triangle from the origin, will equal 2^2 or 4, and statement 1 is thus sufficient.


Statement 2 gives us a point, (sqrt 2, -sqrt2). Again, we can use Pythagorean Theorem to find the distance between that point and the origin, which gives us the radius. sqrt2^2 + sqrt2^2 = 2+2 = 4, so the radius is 2, and that replicates statement 1, which we already know is sufficient.

Thus, both statements are sufficient and the correct answer is D.