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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?

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

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)

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

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)

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

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)

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

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