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Math Help - How do I find the 'Highest' and 'Lowest' point of the Circle

  1. #1
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    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!
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  2. #2
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    Re: How do I find the 'Highest' and 'Lowest' point of the Circle

    Quote Originally Posted by elifast View Post
    r=2
    center= (-1,2)
    (x+1)^2 + (y-2)^2 = 4
    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
    First note the change I made in the equation, =4.

    The high/low points on (x-h)^2+(y-k)^2=r^2 are (h,k+r)~\&~(h,k-r),\,\,r>0.
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  3. #3
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    Re: How do I find the 'Highest' and 'Lowest' point of the Circle

    I'm not sure what liking or not liking "graphing" has to do with this problem. You are told that this is a circle. Use the geometric properties of a circle.

    But if you insist upon doing this "the hard way", you can write (x+1)^2+ (y-2)^2= 4 and differentiate both sides with respect to x: 2(x+1)+ 2(y-2)(dy/dx)= 0. dy/dx= -(x+1)/(y-2)= 0 at a max or min value.
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