My Math Forum Center of Arch and radius

 Algebra Pre-Algebra and Basic Algebra Math Forum

August 28th, 2018, 09:49 PM   #1
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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, 11:25 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1224 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 $r = \sqrt{16+(-3)^2} = \sqrt{25}=5$
 August 29th, 2018, 01:08 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,285 Thanks: 1968 By this theorem, the arch's diameter (twice its radius) is (2 + 4*4/2) ft, so the required radius is 5 ft.
 August 29th, 2018, 08:02 AM #4 Newbie   Joined: Aug 2018 From: NC Posts: 6 Thanks: 0 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.
 August 29th, 2018, 08:10 AM #5 Newbie   Joined: Aug 2018 From: NC Posts: 6 Thanks: 0 I was missing the -4y. That is what threw the whole thing off. I knew it was a simple mistake.

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