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

• Sep 21st 2012, 03:52 AM
elifast
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!
• Sep 21st 2012, 05:03 AM
Plato
Re: How do I find the 'Highest' and 'Lowest' point of the Circle
Quote:

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\$.
• Sep 21st 2012, 06:12 AM
HallsofIvy
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.