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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] New post 09 Aug 2006, 00:17
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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 [#permalink] New post 09 Aug 2006, 04:46
radius = 8
cirumference=16*pi
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 [#permalink] New post 09 Aug 2006, 08:27
I got 16pi as well... wonder if there is a trap somewhere

cir=2pi*radius, so since radius is 8, cir must be 16pi
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 [#permalink] New post 09 Aug 2006, 09:35
Radius = SQRT(9+81) = 3*SQRT(10)

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)
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 [#permalink] New post 09 Aug 2006, 10:13
ps_dahiya wrote:
Radius = SQRT(9+81) = 3*SQRT(10)

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


:shock: did you calculate radius through the triangle and hypotenuse is a radius???
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 [#permalink] New post 09 Aug 2006, 10:17
The radius should be sqrt{(8-1)^2 + (-3)^2} = sqrt (49 + 9) = sqrt(58)

ps_dahiya wrote:
Radius = SQRT(9+81) = 3*SQRT(10)



Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)
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 [#permalink] New post 09 Aug 2006, 10:19
u2lover wrote:
ps_dahiya wrote:
Radius = SQRT(9+81) = 3*SQRT(10)

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


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

My bad.

Radius is distance between point (1,-3) and (8,0)
i.e Radius = SQRT(9+49).
So circumfrence should be = 2 * PI * SQRT(58)
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 [#permalink] New post 09 Aug 2006, 11:42
ps_dahiya wrote:
u2lover wrote:
ps_dahiya wrote:
Radius = SQRT(9+81) = 3*SQRT(10)

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


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

My bad.

Radius is distance between point (1,-3) and (8,0)
i.e Radius = SQRT(9+49).
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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 [#permalink] New post 09 Aug 2006, 11:49
sgrover wrote:
ps_dahiya wrote:
u2lover wrote:
ps_dahiya wrote:
Radius = SQRT(9+81) = 3*SQRT(10)

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


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

My bad.

Radius is distance between point (1,-3) and (8,0)
i.e Radius = SQRT(9+49).
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)

:wall :wall
Misread the question.
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 [#permalink] New post 09 Aug 2006, 11:54
Yeah - its 16*pi :)
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 [#permalink] New post 09 Aug 2006, 13:33
sgrover wrote:
ps_dahiya wrote:
u2lover wrote:
ps_dahiya wrote:
Radius = SQRT(9+81) = 3*SQRT(10)

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


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

My bad.

Radius is distance between point (1,-3) and (8,0)
i.e Radius = SQRT(9+49).
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)


:beat I read the question too quickly... sgrover no wonder you cracked the GMAT !
  [#permalink] 09 Aug 2006, 13:33
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