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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Re: In the xy-plane, point (r, s) lies on a circle with center a [#permalink]
07 Jan 2014, 04:17

Expert's post

7

This post was BOOKMARKED

SOLUTION

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 \(x^2+y^2=r^2\) 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 a [#permalink]
07 Jan 2014, 21:33

If a circle, lying on a xy-plane, has its center at the origin, the equation is x^2+y^2=R^2, where x & y are points on the circle and R is the radius of the circle.

Since x & y from the equation x^2+y^2=R^2 is similar to r & s in the question, we can rewrite as r^2+s^2=R^2.

Statement (1) gives us the value of radius, R; Therefore, r^2+s^2=4; Sufficient. Statement (2) gives us the value of a point on the circle => sub x & y values of the point in r^2+s^2=R^22: 2+2 = 4; r^2 = 4; 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

The formula of circle is given by (x-a)^2 + (y-b)^2 = r^2 where circle is centered as (a,b) and r = radius of the circle.

So if the circle is centered at origin, then the equation reduces to x^2 + y^2 = r^2.

Statement 1) (r,s) lies on the circle and hence r^2 + s^2 should be equal to the radius of the circle and the statement provides that very fact. Hence Sufficient. Statement 2) \((\sqrt{2}, \ -\sqrt{2})\) lies on the circle. Then the radius can be calculated as root(root(2)^2) + root(2)^2) = 2 and as the radius is same from any point on the circle. Hence Sufficient.

Re: In the xy-plane, point (r, s) lies on a circle with center a [#permalink]
11 Jan 2014, 06:14

Expert's post

SOLUTION

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 \(x^2+y^2=r^2\) 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 a [#permalink]
27 Jan 2015, 23:00

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