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

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24 Aug 2010, 22:55

1

This post received KUDOS

Equation of the circle is \(x^2+ y^2 = a^2\) where a = radius

since r,s lies on the circle the equation becomes r^2+s^2 = a^2

statement 1: radius = \(a= 2 => r^2+s^2 = 2^2 = 4\) , thus sufficient. statement 2: we have given one point on the circle and we have the center at the origin. Thus we can find out the distance between these 2 points which will be 2. This distance is basically the radius. Since we have the radius, the statement is sufficient.

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: \((x-a)^2+(y-b)^2=r^2\)

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.

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

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11 Aug 2015, 11:22

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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