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

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Senior Manager
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A circle has the center (1, -3). If the distance between the [#permalink]  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?
Senior Manager
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cirumference=16*pi
VP
Joined: 14 May 2006
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Kudos [?]: 58 [0], given: 0

I got 16pi as well... wonder if there is a trap somewhere

CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Kudos [?]: 119 [0], given: 0

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

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

VP
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Kudos [?]: 58 [0], given: 0

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???
Senior Manager
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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)
CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Kudos [?]: 119 [0], given: 0

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)
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Senior Manager
Joined: 07 Jul 2005
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Location: Sunnyvale, CA
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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)
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 119 [0], given: 0

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)

_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 06 May 2006
Posts: 781
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Kudos [?]: 16 [0], given: 0

Yeah - its 16*pi
Senior Manager
Joined: 14 Jul 2005
Posts: 402
Followers: 1

Kudos [?]: 12 [0], given: 0

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 !
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