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!

Printable View

- Sep 21st 2012, 03:52 AMelifastHow 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 AMPlatoRe: How do I find the 'Highest' and 'Lowest' point of the Circle
- Sep 21st 2012, 06:12 AMHallsofIvyRe: 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.