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getzgetzu
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A circle has the center (1, -3). If the distance between the 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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AK
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 05:46
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radius = 8
cirumference=16*pi
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u2lover
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 09:27
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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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ps_dahiya
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 10:35
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Radius = SQRT(9+81) = 3*SQRT(10)

Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)
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u2lover
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 11:13
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ps_dahiya
Radius = SQRT(9+81) = 3*SQRT(10)

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


:shock: did you calculate radius through the triangle and hypotenuse is a radius???
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gmatornot
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 11:17
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The radius should be sqrt{(8-1)^2 + (-3)^2} = sqrt (49 + 9) = sqrt(58)

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



Circumfrence = 2 * PI * 3 * SQRT(10) = 6 * PI * SQRT(10)
Show more
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 11:19
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u2lover
ps_dahiya
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???
Show more

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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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 12:42
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ps_dahiya
u2lover
ps_dahiya
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)
Show more


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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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 12:49
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sgrover
ps_dahiya
u2lover
ps_dahiya
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)
Show more

:wall :wall
Misread the question.
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 12:54
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Yeah - its 16*pi :)
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gmatornot
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A circle has the center (1, -3). If the distance between the 09 Aug 2006, 14:33
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sgrover
ps_dahiya
u2lover
[quote="ps_dahiya"]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)[/quote]

:beat I read the question too quickly... sgrover no wonder you cracked the GMAT !



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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Thank you for understanding, and happy exploring!