(x - h)^2 + (y - k)^2 = r^2

(x - 5)^2 + (y - 2)^2 = 4^2

If there's a 5 and a 2 to work with, how did 4 come up? I don't understand how to solve these radius distance equations. Can Ipleaseget some help on this?

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- July 22nd 2007, 12:25 PMBlueStarHow did this happen?
(x - h)^2 + (y - k)^2 = r^2

(x - 5)^2 + (y - 2)^2 = 4^2

If there's a 5 and a 2 to work with, how did 4 come up? I don't understand how to solve these radius distance equations. Can I*please*get some help on this? - July 22nd 2007, 12:33 PMJhevon
- July 22nd 2007, 12:37 PMBlueStar
A swimmer jumps 2 feet north and 5 feet east of the corner of the pool. The ripple effect traveled four feet from the center. Model an equation of a circle for the set of points that could be the center of the cannon ball. The corner is the origin at (0,0), and the center is at (5,2) with a radius of 4 feet. I need to find the standard equation using the distance formula.

- July 22nd 2007, 12:56 PMJhevon
as far as i can see, it seems they just want the equation of the circle that forms the cannon ball. in which case it would be an equation of the form (x - h)^2 + (y - k)^2 = r^2 with center (h,k) and radius = 4, which is the standard form of a circle (or do you want to actually derive this equation using the distance formula?)

- July 22nd 2007, 01:23 PMBlueStar
Ohhh, I must have gotten confused with what they were asking. The radius is already answered, obviously.