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

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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 \((\sqrt{2}, \ -\sqrt{2})\) lies on the circle
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Feb 2012, 06:32, edited 2 times in total.
Edited the question and added the OA

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New post 12 Apr 2005, 16:26
saurya_s wrote:
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.

Therefore, (A).
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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:
\((x-a)^2+(y-b)^2=r^2\)


Image

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: \(x^2+y^2=r^2\).

For more on this subject check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

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.
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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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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
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ryusei1989 wrote:
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.


Exactly.
In my opinion it's just poorly formulated as I've seen this exact questioning angle being used as a trap.

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New post 09 Jul 2014, 07:30
mnlsrv wrote:
ryusei1989 wrote:
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.


Exactly.
In my opinion it's just poorly formulated as I've seen this exact questioning angle being used as a trap.


This doubt is addressed here: in-the-xy-plane-point-r-s-lies-on-a-circle-with-center-15566.html#p1191069

Point (r, s) lies ON a circle, means it's on the circumference.
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New post 09 Jul 2015, 02:18
Hi,

i don't understand from Bunuels solution how we come up with the value of S in statement 1 in order to answer what r^2 + s^2 is?

r^2 equals 4, that is all clear, but s should be y-value, how do we get that?

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New post 09 Jul 2015, 03:43
noTh1ng wrote:
Hi,

i don't understand from Bunuels solution how we come up with the value of S in statement 1 in order to answer what r^2 + s^2 is?

r^2 equals 4, that is all clear, but s should be y-value, how do we get that?


\(r^2+s^2=radius^2\). (1) says that radius = 2, thus \(r^2+s^2=2^2\).
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noTh1ng wrote:
Hi,

i don't understand from Bunuels solution how we come up with the value of S in statement 1 in order to answer what r^2 + s^2 is?

r^2 equals 4, that is all clear, but s should be y-value, how do we get that?


Hi noTh1ng,

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
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New post 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!

---------------------------------------------------------------------------------------------------------
Bunuel wrote:
DeeptiM wrote:
OA is D...can anyone explain??


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.

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

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New post 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

Hope this helps.

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New post 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.

Thanks

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New post 03 Mar 2017, 01:40
WilDThiNg wrote:
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.

Thanks


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

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New post 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:
\((x-a)^2+(y-b)^2=r^2\)


Image

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: \(x^2+y^2=r^2\).

For more on this subject check Coordinate Geometry chapter of Math Book: http://gmatclub.com/forum/math-coordina ... 87652.html

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

Therefore, \(r^2+s^2= (\sqrt{2})^2 + ( -\sqrt{2})^2\) = 2+2 = 4

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New post 03 Mar 2017, 08:20
Bunuel: thanks for clarifying!

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New post 20 Jul 2017, 23:09
So, for statement 2, any coordinates that are on the circle can be used to substitute for (r,s)?

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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 \(x^2+y^2=radius^2\)
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New post 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?

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New post 21 Jul 2017, 00:26
parkerd wrote:
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?
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Re: In the xy-plane, point (r, s) lies on a circle with center   [#permalink] 21 Jul 2017, 00:26

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

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