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25 Aug 2010, 04:46
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C
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Circle C is in the xy-plane, what is the area of the circle?
(1) Points (-2, 0) and (0,2) lie on the circle.
(2) The radius of the circle is equal to or less than 2^(1/2).
(1) Points (-2, 0) and (0,2) lie on the circle.
(2) The radius of the circle is equal to or less than 2^(1/2).
25 Aug 2010, 05:29
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jitendra
\(area=\pi{r^2}\), so we should find the value of radius.
It would be better if you visualize this problem.
(1) Points (-2, 0) and (0,2) lie on the circle --> two points DO NOT define a circle (three points does), hence we can have numerous circles containing these two points, thus we can not find single numerical value of radius. Not sufficient.
Side note: if you put points (-2, 0) and (0,2) on XY-plane you can see that center of the circle must be on the line \(y=-x\) (the center of the circle must be equidistant from two pints given).
(2) The radius of the circle is equal to or less than \(\sqrt{2}\) --> \(r\leq{\sqrt{2}}\). Clearly insufficient.
(1)+(2) The distance between the 2 points given is \(d=\sqrt{2^2+2^2}=2\sqrt{2}\), so it's min length of diameter of the circle passing these points (diameter of a circle passing 2 points can not be less than the distance between these 2 points), thus half of \(2\sqrt{2}\) is min length of the radius of the circle --> \(r\geq{\sqrt{2}}\) but as from (2) \(r\leq{\sqrt{2}}\) then \(r=\sqrt{2}\) --> \(area=\pi{r^2}=2\pi\). Sufficient.
Answer: C.
General Discussion
25 Aug 2010, 10:03
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Good question...As Bunuel said its easier to visualise this and solve than just solving on paper!
