My Math Forum Center of the circle calculation

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 June 25th, 2016, 09:52 AM #1 Newbie   Joined: Jun 2016 From: Austin tx Posts: 2 Thanks: 0 Center of the circle calculation I've been at this for an hour now and I can't figure out how to come to the correct answer J. How would one arrive at this answer?
 June 25th, 2016, 10:19 AM #2 Senior Member     Joined: Feb 2010 Posts: 703 Thanks: 138 The given point $\displaystyle (\sqrt{3},0)$ is not needed. Let $\displaystyle P(x,y)$ be any point on the circle. You are given $\displaystyle PB=2PO$. Use the distance formula, square, expand, and put into standard form. You will end up with $\displaystyle x^2+(y+1)^2=4$ which tells you the center is $\displaystyle (0,-1)$.
 June 25th, 2016, 01:36 PM #3 Newbie   Joined: Jun 2016 From: Austin tx Posts: 2 Thanks: 0 How did you arrive at the standard form $\displaystyle x^2+(y+1)^2=4$ using the distance formula?
June 25th, 2016, 02:30 PM   #4
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 Originally Posted by herzhand How did you arrive at the standard form $\displaystyle x^2+(y+1)^2=4$ using the distance formula?
$\displaystyle PB = \sqrt{(x-0)^2 + (y-3)^2}$

$\displaystyle PO = \sqrt{(x-0)^2 + (y-0)^2}$

 July 8th, 2016, 04:14 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 And you are told that PB= 2PO.

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