Pre-Calculus Pre-Calculus Math Forum

 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
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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. Tags calculation, center, circle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jiokidx Geometry 1 June 18th, 2015 08:20 AM fornuftit Algebra 2 May 23rd, 2012 10:23 PM Singularity Complex Analysis 3 November 11th, 2010 01:59 AM thyrgle Algebra 4 May 16th, 2010 04:10 PM rune2402 Algebra 4 February 27th, 2008 05:21 PM

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