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June 25th, 2016, 10:52 AM   #1
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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?
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June 25th, 2016, 11:19 AM   #2
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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 $\displaystyle (0,-1)$.
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June 25th, 2016, 02:36 PM   #3
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How did you arrive at the standard form $\displaystyle x^2+(y+1)^2=4$ using the distance formula?
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June 25th, 2016, 03:30 PM   #4
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Quote:
Originally Posted by herzhand View Post
How did you arrive at the standard form $\displaystyle x^2+(y+1)^2=4$ using the distance formula?
$\displaystyle PB = \sqrt{(x-0)^2 + (y-3)^2}$


$\displaystyle PO = \sqrt{(x-0)^2 + (y-0)^2}$
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July 8th, 2016, 05:14 AM   #5
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And you are told that PB= 2PO.
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