# Thread: How do I find the 'Highest' and 'Lowest' point of the Circle

1. ## 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!

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

Originally Posted by elifast
r=2
center= (-1,2)
$\displaystyle (x+1)^2 + (y-2)^2 = 4$
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, $\displaystyle =4$.

The high/low points on $\displaystyle (x-h)^2+(y-k)^2=r^2$ are $\displaystyle (h,k+r)~\&~(h,k-r),\,\,r>0$.

3. ## 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 $\displaystyle (x+1)^2+ (y-2)^2= 4$ 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.