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In the xyplane, point (r, s) lies on a circle with center a
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07 Jan 2014, 05:17
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The Official Guide For GMAT® Quantitative Review, 2ND EditionIn the xyplane, 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 Data Sufficiency Question: 22 Category: Geometry Simple coordinate geometry Page: 154 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach 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. Thank you!
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Re: In the xyplane, point (r, s) lies on a circle with center a
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07 Jan 2014, 05:17
SOLUTIONTHEORY:In an xyplane, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\) 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 rightangled triangle whose other sides are of length xa and yb. 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: mathcoordinategeometry87652.htmlBACK TO THE ORIGINAL QUESTION:In the xyplane, 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 xyplane, point (r, s) lies on a circle with center a
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07 Jan 2014, 22:33
If a circle, lying on a xyplane, 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.
Hence, Answer is (D).



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Re: In the xyplane, point (r, s) lies on a circle with center a
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08 Jan 2014, 09:04
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIn the xyplane, 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 (xa)^2 + (yb)^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. Option D)
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Re: In the xyplane, point (r, s) lies on a circle with center a
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11 Jan 2014, 07:14
SOLUTIONTHEORY:In an xyplane, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\) 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 rightangled triangle whose other sides are of length xa and yb. 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: mathcoordinategeometry87652.htmlBACK TO THE ORIGINAL QUESTION:In the xyplane, 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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: In the xyplane, point (r, s) lies on a circle with center a
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09 Oct 2016, 02:09
So laying on circle means on the edge of the circle? could not mean inside the circle? Someone please clarify.



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Re: In the xyplane, point (r, s) lies on a circle with center a
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09 Oct 2016, 02:23
TheLordCommander wrote: So laying on circle means on the edge of the circle? could not mean inside the circle? Someone please clarify. Yes, "on a circle " means on the edge only. Had the point been inside the circle, it would have been clearly mentioned.
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Re: In the xyplane, point (r, s) lies on a circle with center a
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