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Point on the perimeter.

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by okletsdothis » Sat Feb 26, 2011 5:12 am
How do you solve. Please explain

A circle is mapped on the x,y coordinate plane, with the center of the circle at the origin. If point A is located on
the perimeter of the circle, what is the sum of the squares of its x and y coordinates?

(1) The radius of the circle is 2.
(2) One of the points on the perimeter of the circle is (√2, √2).

OA D
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Source: — Data Sufficiency |
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by anshumishra » Sat Feb 26, 2011 5:28 am
okletsdothis wrote:How do you solve. Please explain

A circle is mapped on the x,y coordinate plane, with the center of the circle at the origin. If point A is located on
the perimeter of the circle, what is the sum of the squares of its x and y coordinates?

(1) The radius of the circle is 2.
(2) One of the points on the perimeter of the circle is (√2, √2).

OA D
Say O is the center [at origin i.e at (0,0)] of the circle and any point A (x,y) lies on the perimeter.
x^2+y^2 = ?

The distance between A(x,y) and O(0,0) is nothing but the radius of the circle
=> √(x^2+y^2) = radius of the circle ---- 1

Statement 1
The radius of the circle is 2

Clearly, using this and (1) as shown above we can find x^2+y^2 = radius ^2 ----- Sufficient

Statement 2:
One of the points on the perimeter of the circle is (√2, √2).

Since any point on the circumference of the circle is equal to radius of the circle, so again we know the radius of the circle =
√(x^2+y^2) = √4 (please note you don't need to calculate this in real exam, I am showing it here to explain only).
So, again x^2+y^2 = 2 ---- Sufficient

Hence, D
Last edited by anshumishra on Sat Feb 26, 2011 5:49 am, edited 1 time in total.
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Anshu

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by okletsdothis » Sat Feb 26, 2011 5:38 am
See what my doubt is... will 'sum of the squares of its x and y coordinates' always be the same value ?

I know how to get tht value for a particular set of numbers, but all the x y co-ordinates on the circumference when squared and added will have the same value is something that i am not sure your explanation covers properly.

There will be many points on the circumference which will be a peculiar number with different decimal points and when you square them will you always get this particular number ?

How am I suppose to know this ?
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by anshumishra » Sat Feb 26, 2011 5:46 am
okletsdothis wrote:See what my doubt is... will 'sum of the squares of its x and y coordinates' always be the same value ?

I know how to get tht value for a particular set of numbers, but all the x y co-ordinates on the circumference when squared and added will have the same value is something that i am not sure your explanation covers properly.

There will be many points on the circumference which will be a peculiar number with different decimal points and when you square them will you always get this particular number ?

How am I suppose to know this ?
Yes, it will be the same.
Since the point A (x,y) lies on the circumference of the circle. It's distance from the origin O (which is the same thing sqrt(x^2+y^2) as the radius. You have been given the radius in both the statements. So both are sufficient.

Here you can see that if point (x,y) lies any where on the unit circle (with radius 1 not 2 as is the case here though), the value of x^2+y^2 would be equal : https://en.wikipedia.org/wiki/File:Unit_ ... _color.svg
Thanks
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by Anurag@Gurome » Sat Feb 26, 2011 8:53 pm
I got a PM to reply on this thread. So, here it goes. See the attachment, I hope that may help.

(1) Since OA = 2 = OA' = OA'' (I have taken the point A' and A'' for clarification of your query)
So, OA = √(x^2 + y^2) = 2 (By distance formula)
So, x^2 + y^2 = 2^2 = 4
So, (1) is SUFFICIENT.

(2) One of the points on circle is (√2, √2) and (0, 0) is the center of circle.
So, radius = √{(√2 - 0)^2 + (√2 - 0)^2)} = √(2 + 2) = 2
So, again radius of circle = 2, so from it can followed as we did in statement 1.
So, (2) is SUFFICIENT.

The correct answer is D.
Attachments
Circle.doc
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Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

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by okletsdothis » Sat Feb 26, 2011 10:31 pm
Thanks anshumishra and Anurag.
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