Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In the xy-plane, point (r, s) lies on a circle with center 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.

(1) r^2 + s^2 is the square of the radius of the circle. Sufficient.

(2) This is of no consequence since for any circle centered at the origin, there would be a point (v2. -v2) would lie on the circle. Gives us no info about r^2 + s^2.

THEORY: In an xy-plane, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((x-a)^2+(y-b)^2=r^2\)

This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\).

BACK TO THE ORIGINAL QUESTION: In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of \(r^2 + s^2\)?

Now, as \(x^2+y^2=radius^2\) then the question asks about the value of radius^2.

(1) The circle has radius 2 --> radius^2=4. Sufficient.

(2) The point \((\sqrt{2}, \ -\sqrt{2})\) lies on the circle --> substitute x and y coordinates of a point in \(x^2+y^2=radius^2\) --> \(2+2=4=r^2\). Sufficient.

Re: In the xy-plane, point (r, s) lies on a circle with center [#permalink]

Show Tags

02 Mar 2013, 22:08

2

This post received KUDOS

Thanks for the brilliant explanation. One thing I don't get the question is that, the point (r,s) could be anywhere in the circle, not only on its circumference. Why does it refer only to a point on the circumference? Thanks!

Re: In the xy-plane, point (r, s) lies on a circle with center [#permalink]

Show Tags

03 Mar 2013, 00:02

2

This post received KUDOS

ryusei1989 wrote:

Thanks for the brilliant explanation. One thing I don't get the question is that, the point (r,s) could be anywhere in the circle, not only on its circumference. Why does it refer only to a point on the circumference? Thanks!

It is the language. On the circle = On the circumference. In/Inside/Within the circle = Points enclosed by the circumference
_________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks!

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

(1) The circle has radius 2 (2) The point (2√, −2√) lies on the circle

You seem to have misunderstood a little here.

The equation of Circle is given by \(x^2 + y^2 = Radius^2\)

Given : (r,s) lie on the circle i.e. (r,s) will satisfy the equation of Circle i.e. \(r^2 + s^2 = Radius^2\)

Question : Find the value of \(r^2 + s^2\)? but since \(r^2 + s^2 = Radius^2\) therefore, the question becomes

Question : Find the value of \(Radius^2\)?

Statement 1: The circle has radius 2 i.e. \(r^2 + s^2 = Radius^2 = 2^2 = 4\) SUFFICIENT

Statement 2: The point (√2, −√2) lies on the circle i.e. (√2, −√2) will satisfy the equation of circle i.e. (√2)^2 + (−√2)^2 = Radius^2 i.e. Radius = 4 hence, \(r^2 + s^2 = Radius^2 = 2^2 = 4\) Hence, SUFFICIENT

Answer: Option D

I hope it helps!

Please Note: You have been confused r (X-co-ordinate) and r (Radius) as it seems from your question _________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

In the xy-plane, point (r, s) lies on a circle with center [#permalink]

Show Tags

07 Jan 2016, 12:01

I'm quoting Bunuel's correct explanation below with some follow up questions, since this question's wording did not seem precise to me:

1. The given information says that point (r,s) lies on the circle. For future reference, will "on" mean on the circumference of the circle? In other words, "on" can never refer to a point inside the circle. If the point (r,s) were some arbitrary point inside of the circle, the solution to this problem would be incorrect I believe.

2. Is the point (r,s) assumed to NOT be constant? It seems like the explanation for working out (2) depends upon the fact that (r,s) is non-constant and could be ANY point along the circumference of the circle. If (r,s) were in fact constant, then I don't believe (2) would provide any information in deducing what r^2+s^2 is.

EDIT: I missed the above clarification regarding the language for "on the circle". I have crossed that question of mine out. Sorry!

BACK TO THE ORIGINAL QUESTION: In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of \(r^2 + s^2\)?

Now, as \(x^2+y^2=radius^2\) then the question asks about the value of radius^2.

(1) The circle has radius 2 --> radius^2=4. Sufficient.

(2) The point \((\sqrt{2}, \ -\sqrt{2})\) lies on the circle --> substitute x and y coordinates of a point in \(x^2+y^2=radius^2\) --> \(2+2=4=r^2\). Sufficient.

Re: In the xy-plane, point (r, s) lies on a circle with center [#permalink]

Show Tags

07 Jan 2016, 12:44

lillylw wrote:

I'm quoting Bunuel's correct explanation below with some follow up questions, since this question's wording did not seem precise to me:

1. The given information says that point (r,s) lies on the circle. For future reference, will "on" mean on the circumference of the circle? In other words, "on" can never refer to a point inside the circle. If the point (r,s) were some arbitrary point inside of the circle, the solution to this problem would be incorrect I believe.

2. Is the point (r,s) assumed to NOT be constant? It seems like the explanation for working out (2) depends upon the fact that (r,s) is non-constant and could be ANY point along the circumference of the circle. If (r,s) were in fact constant, then I don't believe (2) would provide any information in deducing what r^2+s^2 is.

EDIT: I missed the above clarification regarding the language for "on the circle". I have crossed that question of mine out. Sorry!

For analysing statement 2, it does not matter whether (r,s) is constant.

Equation of a circle with center (a,b) and radius R is \((x-a)^2+(y-b)^2=R^2\) , now as the given circle is centered at (0,0) ---> a=b=0 ---> the equation of the circle thus becomes \(x^2+y^2 = R^2\)

As (r,s) lies on the circle ---> you can substitute r for x and s for y in the equation of the circle to get, \(r^2+s^2=R^2\)

Per statement 2, [\(\sqrt{2} , -\sqrt{2}\)] lies on the circle ---> from equation of the circle \(x^2+y^2 = R^2\) ---> \((\sqrt{2})^2+(-\sqrt{2})^2=R^2\)

But as mentioned above, \(r^2+s^2=R^2\) ---> \(r^2+s^2 =(\sqrt{2})^2+(-\sqrt{2})^2 = 4\) = a unique value.

Thus, it does not matter what particular values r and s take. Whatever set of (r,s) values we will get, will always satisfy the equation \(r^2+s^2=4\). Some of the sets can be

\((\sqrt{2}), -\sqrt{2})\) or \((1, -\sqrt{3})\) or \((1, \sqrt{3})\) or \((-\sqrt{3},1)\) or \((\sqrt{3}, 1)\) etc

Re: In the xy-plane, point (r, s) lies on a circle with center [#permalink]

Show Tags

02 Mar 2017, 21:12

Hi,

A very fundamental or, maybe, a silly question, in GMAT, when a question like this reads 'on a circle', it's supposed to mean, on the circumference of the circle, and not the whole of the area of the circle.

A very fundamental or, maybe, a silly question, in GMAT, when a question like this reads 'on a circle', it's supposed to mean, on the circumference of the circle, and not the whole of the area of the circle.

Thanks

Yes, on the circle means on the circumference. In the circle means within.
_________________

Re: In the xy-plane, point (r, s) lies on a circle with center [#permalink]

Show Tags

03 Mar 2017, 01:51

Bunuel wrote:

DeeptiM wrote:

OA is D...can anyone explain??

THEORY: In an xy-plane, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((x-a)^2+(y-b)^2=r^2\)

This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\).

BACK TO THE ORIGINAL QUESTION: In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of \(r^2 + s^2\)?

Now, as \(x^2+y^2=radius^2\) then the question asks about the value of radius^2.

(1) The circle has radius 2 --> radius^2=4. Sufficient.

(2) The point \((\sqrt{2}, \ -\sqrt{2})\) lies on the circle --> substitute x and y coordinates of a point in \(x^2+y^2=radius^2\) --> \(2+2=4=r^2\). Sufficient.

Answer: D.

Hope it helps.

Another way of using St II is since (r,s) and \((\sqrt{2}, \ -\sqrt{2})\) lie on the circle Their distance from the center(origin in this case) will be same

So, for statement 2, any coordinates that are on the circle can be used to substitute for (r,s)?

Yes, if a circle is centred at the origin, then the x and y-coordinates of any point on the circle (so on the circumference) will satisfy \(x^2+y^2=radius^2\)
_________________

Re: In the xy-plane, point (r, s) lies on a circle with center [#permalink]

Show Tags

21 Jul 2017, 00:09

If the (r,s) was inside the circle would you still be able to solve with the same equation like in statement 1? Or be able to substitute for (r,s) with the given values on the circle like in statement 2?

If the (r,s) was inside the circle would you still be able to solve with the same equation like in statement 1? Or be able to substitute for (r,s) with the given values on the circle like in statement 2?

If (r, s) were IN the circle, then the answer would be E because each statement gives us basically the same info - the length of the radius. How can we find the sum of the squares of a random point inside the circle just knowing the radius?
_________________