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 Post subject: GMAT Coordinate Geometry
PostPosted: Sat May 04, 2013 7:02 am 
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A circle is drawn on the 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).

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.


(D) Statement (1) is sufficient by itself. For a circle with its center at the origin of the coordinate plane, the value of the sum of the squares of the coordinates of the points on the perimeter can be determined. This value is simply the square of the radius of the circle (2² = 4).
For example, one (x, y) coordinate that is on the perimeter of the circle is (2, 0). The sum of the squares of x and y would equal 4. Likewise, using the Pythagorean Theorem, for all points (x, y) on the circle, x² + y² = 4.

Statement (2) is also sufficient. The sum of the squares of the x and y coordinates given in Statement (2) is: (-2)² + (-2)² = 4.

Since both statements are sufficient individually, the correct answer is choice (D).
-------------

Please, kindly explain the statement quoted from your explanation below:

"Likewise, using the Pythagorean Theorem, for all points (x, y) on the circle, x² + y² = 4."


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 Post subject: Re: GMAT Coordinate Geometry
PostPosted: Sat May 04, 2013 7:03 am 
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For any point C (x, y) which lies on the circle but not on the axis, we can build a triangle ABC (see graphics).

When we apply Pythagorean Theorem to this triangle we will get:
AC² = AB² + BC²
2² = |x|² + |y
4 = x² + y²
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