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August 28th, 2018, 08:49 PM  #1 
Newbie Joined: Aug 2018 From: NC Posts: 6 Thanks: 0  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

August 28th, 2018, 10:25 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 
It should be pretty clear that the xcoordinate of the circle center is $x=0$ Let $y$ be the ycoordinate 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... $16+y^2 = (2y)^2$ $16+y^2 = 4  4y+ y^2$ $12 = 4y$ $y=3$ 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, 12:08 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,542 Thanks: 1750 
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, 07:02 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, 07:10 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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