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In the rectangular coordinate system, points (4, 0) and ( 4,

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VP
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In the rectangular coordinate system, points (4, 0) and ( 4, [#permalink] New post 08 Jul 2008, 19:17
00:00
A
B
C
D
E

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In the rectangular coordinate system, points (4, 0) and (– 4, 0) both lie on circle C. What is the
maximum possible value of the radius of C ?
A. 2
B. 4
C. 8
D. 16
E. There is no finite maximum value.

I know the OA.. So please dont post A,B..type answers.Please explain your work.
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Re: PS-Circle [#permalink] New post 08 Jul 2008, 19:25
goalsnr wrote:
In the rectangular coordinate system, points (4, 0) and (– 4, 0) both lie on circle C. What is the
maximum possible value of the radius of C ?
A. 2
B. 4
C. 8
D. 16
E. There is no finite maximum value.

I know the OA.. So please dont post A,B..type answers.Please explain your work.


Here is how I would've solved on the actual exam. Since you know that the center lies anywhere on y-axis -> E is the answer

Trying to come up with explanation. One - geometrically - draw a triangle AOB, AO = BO, etc
Analytically: (a,b) center of the circle. (x-a)^2 + (y-b)^2 = R^2

(4-a)^2 + (0-b)^2 = R^2
(-4-a)^2 + (0-b)^ = R^2

subtract those two. 16*a = 0 -> a = 0
put it back: R^2 = 16 + b^2

Easy to see that for any b, there is a circle x^2 + (y-b)^2 = b^2+16 that goes through both (-4,0) and (4,0) -> E
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Re: PS-Circle [#permalink] New post 08 Jul 2008, 19:28
is it E.

All we know is length of a line segament of a circle. 8
there could be infinte number of circles with different sizes and can have the same line segament.
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Re: PS-Circle [#permalink] New post 08 Jul 2008, 20:01
Thanks Guys.

Oa Is E
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Re: PS-Circle [#permalink] New post 09 Jul 2008, 03:39
Just notice that any circle whose center would be equidistant from the two points would work and that the locus of the equidistant points is the perpendicular bisector of the segment between those two points.

It is an infinite line ==> answer is (E)
Re: PS-Circle   [#permalink] 09 Jul 2008, 03:39
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In the rectangular coordinate system, points (4, 0) and ( 4,

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