My Math Forum How do I find the 'Highest' and 'Lowest' point of the Circle

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 September 21st, 2012, 03:52 AM #1 Newbie   Joined: Sep 2012 Posts: 15 Thanks: 0 How do I find the 'Highest' and 'Lowest' point of the Circle r=2 center= (-1,2) (x+1)^2 + (y-2)^2 = 2 I was thinking I just take the radius and add it + or - on the Y axis from the center? Is there any analytical way to do this? I really don't like graphing! Thank a lot!
 September 21st, 2012, 04:15 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: How do I find the 'Highest' and 'Lowest' point of the Ci Given the circle: $(x-h)^2+(y-k)^2=r^2$ The points you desire are: $\(h,k\pm r)$ In other words, the "highest" point is r units directly above the center and the "lowest point is r units directly below the center.

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# what is lowest point of a circle

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