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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: 706 Thanks: 141 
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 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 141  
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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