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# A circle has the center (1, -3). If the distance between the

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A circle has the center (1, -3). If the distance between the [#permalink]

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09 Aug 2006, 01:17
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A circle has the center (1, -3). If the distance between the center and one of the intersections with x-axis is 8. What is the circumfrence of the circle?
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09 Aug 2006, 05:46
cirumference=16*pi
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09 Aug 2006, 09:27
I got 16pi as well... wonder if there is a trap somewhere

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09 Aug 2006, 10:35

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)
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09 Aug 2006, 11:13
ps_dahiya wrote:

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)

did you calculate radius through the triangle and hypotenuse is a radius???
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09 Aug 2006, 11:17
The radius should be sqrt{(8-1)^2 + (-3)^2} = sqrt (49 + 9) = sqrt(58)

ps_dahiya wrote:

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)
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09 Aug 2006, 11:19
u2lover wrote:
ps_dahiya wrote:

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)

did you calculate radius through the triangle and hypotenuse is a radius???

Radius is distance between point (1,-3) and (8,0)
So circumfrence should be = 2 * PI * SQRT(58)
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09 Aug 2006, 12:42
ps_dahiya wrote:
u2lover wrote:
ps_dahiya wrote:

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)

did you calculate radius through the triangle and hypotenuse is a radius???

Radius is distance between point (1,-3) and (8,0)
So circumfrence should be = 2 * PI * SQRT(58)

I am confused. The question no where mentions that it intersects at (8,0) . It clearly mentions that the distance between center and point on circle is 8. The radius of the circle shd be 8, and circumfrence = 16*pi

The point onthe X-axis hten shd be: ((sqrt (64-9) + 1), 0)
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09 Aug 2006, 12:49
sgrover wrote:
ps_dahiya wrote:
u2lover wrote:
ps_dahiya wrote:

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)

did you calculate radius through the triangle and hypotenuse is a radius???

Radius is distance between point (1,-3) and (8,0)
So circumfrence should be = 2 * PI * SQRT(58)

I am confused. The question no where mentions that it intersects at (8,0) . It clearly mentions that the distance between center and point on circle is 8. The radius of the circle shd be 8, and circumfrence = 16*pi

The point onthe X-axis hten shd be: ((sqrt (64-9) + 1), 0)

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09 Aug 2006, 12:54
Yeah - its 16*pi
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Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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09 Aug 2006, 14:33
sgrover wrote:
ps_dahiya wrote:
u2lover wrote:
ps_dahiya wrote:

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)

did you calculate radius through the triangle and hypotenuse is a radius???

Radius is distance between point (1,-3) and (8,0)
So circumfrence should be = 2 * PI * SQRT(58)

I am confused. The question no where mentions that it intersects at (8,0) . It clearly mentions that the distance between center and point on circle is 8. The radius of the circle shd be 8, and circumfrence = 16*pi

The point onthe X-axis hten shd be: ((sqrt (64-9) + 1), 0)

I read the question too quickly... sgrover no wonder you cracked the GMAT !
09 Aug 2006, 14:33
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