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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 [#permalink] New post 09 Dec 2010, 09:31
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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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Re: Point on a circle [#permalink] New post 09 Dec 2010, 09:45
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udaymathapati 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 (\sqrt{\(2\)}, -\sqrt{\(2\)}) lies on the circle.


THEORY:
In an x-y Cartesian coordinate system, 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:
\(x^2+y^2=r^2\)

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 \(x^2+y^2=r^2\) 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 \(x^2+y^2=r^2\) --> \(2+2=4=r^2\). Sufficient.

Answer: D.
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Re: In the xy-plane, point (r, s) lies on a circle with center [#permalink] New post 11 Aug 2015, 10:22
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Re: In the xy-plane, point (r, s) lies on a circle with center   [#permalink] 11 Aug 2015, 10:22
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