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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).

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.

Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]
28 Jun 2013, 21:46

1

This post received KUDOS

Adding to Bunnel's explanation, another way to arrive at why r >=\sqrt{2} is because the sum of any two sides of a triangle is greater than or equal to the third side. The triangle under consideration would be the one formed by the points (-2,0) , (0,2) and the center of the circle. Two sides of this triangle are equal to the radii of the circle. Thus, r+r >= distance between points (-2,0) , (0,2) = 2 \sqrt{2}. Thanks.

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).

Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]
08 Mar 2015, 04:52

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