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August 28th, 2018, 08:49 PM   #1
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Center of Arch and radius

Matt is remodeling the front entrance to his home and needs to cut an arch for the top of an entranceway. The arch must be 8 ft wd and 2 ft high. To draw the arch, he will use a stretched string with chalk attached at an end as a compass. a) using the coordinate system, located the center of the circle. b) what radius should Matt use to draw the arch? Use of distance formula is in effect but I really can't solve this one
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August 28th, 2018, 10:25 PM   #2
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It should be pretty clear that the x-coordinate of the circle center is $x=0$

Let $y$ be the y-coordinate of the center.

The point $(0,y)$ is equidistant from the points $(-4,0),~(4,0),~(0,2)$ as this is the common radius of the circle. As the distances are the same the squared distances will be as well. So...

$16+y^2 = (2-y)^2$

$16+y^2 = 4 - 4y+ y^2$

$12 = -4y$

$y=-3$

The radius is just the square root of either side of the above equation

$r = \sqrt{16+(-3)^2} = \sqrt{25}=5$
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August 29th, 2018, 12:08 AM   #3
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By this theorem, the arch's diameter (twice its radius) is (2 + 4*4/2) ft, so the required radius is 5 ft.
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August 29th, 2018, 07:02 AM   #4
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Thank you!! I tried several ways to figure this one out and could not get anywhere! Now I'll know how to do it if this type of problem pops up again.
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August 29th, 2018, 07:10 AM   #5
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I was missing the -4y. That is what threw the whole thing off. I knew it was a simple mistake.
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