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alimad
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Yurik79
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alimad
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Hmm.. I'm getting something completely different.

Area of a circle = (pi)r^2

If the x value is 4, and the circle has to remain positive, then the radius can at most be 4.

So, (pi)4^2 = (pi)16 = ~496
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cpetro29
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16pi

If the center is 4,7, the largest radius, with everything being positive, would be 4. If it were 7, then one of the points would be -3,7, which isn't positive.
I get -3,7 by taking the center (4,7) and moving to the left 7 points (your radius option).

Hope this helps.
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GMATT73
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cpetro29
16pi

If the center is 4,7, the largest radius, with everything being positive, would be 4. If it were 7, then one of the points would be -3,7, which isn't positive.
I get -3,7 by taking the center (4,7) and moving to the left 7 points (your radius option).

Hope this helps.


Used the exact same method. A radius of 7 would sweep into the II quadrant. To keep (a,b) coordinates in the Ist quad, we must use 4 as the radius of the circle..

16*pi
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alimad
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excellent explaination. Thank you gentleman
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Yurik79
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cpetro29
16pi

If the center is 4,7, the largest radius, with everything being positive, would be 4. If it were 7, then one of the points would be -3,7, which isn't positive.
I get -3,7 by taking the center (4,7) and moving to the left 7 points (your radius option).

Hope this helps.

Nothing to add Gr8 explanation
alimad just draw a schematic picture and you ll see it
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haas_mba07
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Answer: 16 x pi


For all co-ordinates ON the circle to be positive, the closest the circle can be is touching the two axis. In that case, x-coordinate = y-coordinate for the center.

We know (x, y) = (4,7), which means the circle is touching the y axis and not the x. Radius = x co-ordinate.

Therefore, area = pi x 4 x 4 = pi x 16.



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