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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.

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

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11 Feb 2012, 06:30

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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.

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

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02 Mar 2013, 22:08

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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!

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03 Mar 2013, 00:02

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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!

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05 Jul 2014, 02:53

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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!

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

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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]

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07 Jan 2016, 12:44

Expert's post

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

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

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07 Jan 2016, 12:46

Quote:

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

Hope this helps.

Perfect thank you! I understand this now!

gmatclubot

In the xy-plane, point (r, s) lies on a circle with center
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07 Jan 2016, 12:46

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