In the xy plane, point (r,s) lies on the circle with the cen

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mitzwillrockgmat: Master | Next Rank: 500 Posts; Posts: 124; Joined: Wed May 19, 2010 10:20 pm; Thanked: 3 times; GMAT Score:1100

by mitzwillrockgmat » Sat Jun 12, 2010 2:20 pm

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

1. The circle has radius 2
2. The point (v2,-v2) lies on the circle v=square root

Help out with statement 2 please!

1. radius is 2, so i can use circle of eq formula x^2 +y^2= R^2 (radius)... so r^2 +s^2 =2^2 => 4

2. im a bit confused here - does it simply mean that we can substitute sqroot 2 & (-) sqroot 2 to get answer? so then, (sqroot2)^2 + [ - (sqroot 2)]^2 = r^2 +s^2 ? hence, the answer agn is 4

I'm thinking that this must be correct since answer is D.

By the way i could've also interpreted "on the circle" to mean that the points are inside the circle.
So, is it safe to say that "on the circle" ALWAYS means that the points are on the circumference & NOT, inside the circle?

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Source: Beat The GMAT — Data Sufficiency | Sat Jun 12, 2010 2:20 pm

Rich@VeritasPrep: GMAT Instructor; Posts: 147; Joined: Tue Aug 25, 2009 7:57 pmFollowed by: 17 members; Location: New York City; Thanked: 76 times; GMAT Score:770

by Rich@VeritasPrep » Sat Jun 12, 2010 2:27 pm

Mitz,

Yes, you are correct.

If the point (sqrt(2), -sqrt(2)) is on the circle, that means that the x and y coordinates both have the same absolute value, which means that you could draw a 45-45-90 right triangle with the points (0,0), (sqrt(2), 0), and (sqrt(2), -sqrt(2)) as vertices.

The hypotenuse of this triangle, which has length 2, would also be the radius of the circle. r^2+s^2 is the square of the radius, and thus you can find r^2 + s^2.

Also, "on the circle" will always mean along the circumference. If the question means to specify a point inside the circle, it will state as exactly that or something similar, such as "within the circle".

Rich Zwelling
GMAT Instructor, Veritas Prep

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mitzwillrockgmat: Master | Next Rank: 500 Posts; Posts: 124; Joined: Wed May 19, 2010 10:20 pm; Thanked: 3 times; GMAT Score:1100

by mitzwillrockgmat » Sat Jun 12, 2010 2:44 pm

raz1024 wrote:Mitz,

Yes, you are correct.

If the point (sqrt(2), -sqrt(2)) is on the circle, that means that the x and y coordinates both have the same absolute value, which means that you could draw a 45-45-90 right triangle with the points (0,0), (sqrt(2), 0), and (sqrt(2), -sqrt(2)) as vertices.

The hypotenuse of this triangle, which has length 2, would also be the radius of the circle. r^2+s^2 is the square of the radius, and thus you can find r^2 + s^2.

Also, "on the circle" will always mean along the circumference. If the question means to specify a point inside the circle, it will state as exactly that or something similar, such as "within the circle".

Cool! thanks Raz

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8183; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: In the xy plane, point (r,s) lies on the circle with the cen

by Scott@TargetTestPrep » Sun Jul 04, 2021 3:50 am

mitzwillrockgmat wrote: ↑
Sat Jun 12, 2010 2:20 pm
In the xy plane, point (r,s) lies on the circle with the centre at the origin. What is the value of r^2 +s^2?

1. The circle has radius 2
2. The point (v2,-v2) lies on the circle v=square root

Solution:

Question Stem Analysis:

We need to determine the value of r^2 + s^2, given that (r, s) is a point on the circle with center at the origin. Notice that the equation of such a circle is x^2 + y^2 = R^2 where R is the radius of the circle. Since (r, s) is a point on the circle, we have r^2 + s^2 = R^2. That is, in order to determine the value of r^2 + s^2, we either need to know the coordinates of a point on the circle or just the value of R.

Statement One Alone:

Since the radius of the circle is 2, R = 2. Therefore, r^2 + s^2 = 2^2 = 4. Statement one alone is sufficient.

Statement Two Alone:

Since (√2, -√2) is a point on the circle, x = √2 and y = -√2 will satisfy x^2 + y^2 = R^2. Substituting, we find R^2 = x^2 + y^2 = (√2)^2 + (-√2)^2 = 2+ 2 = 4. Proceeding as above, we obtain r^2 + s^2 = 4. Statement two alone is sufficient.

Answer: D

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