equation of circle w/ diam. endpoints at (-3,-2) and (1,4

[bubblefiz]

I was wondering if you could check this to see if it looks right and if not suggest any help.

Find an equation of a circle that contains the end points of a diameter at (-3, -2) and (1, 4).

First I graphed these to get a center of (h,k) = (-1,1)

Then to get radius (I'm not sure if this part is right)

A^2+B^2=C^2
(2)^2+(3)^2=c^2
4 + 9 = C^2
Radius is squareroot 13

so, then
(x-(-1))^2+ (y-(1))^2=(13)^2
(x+1)^2 + (y-1)^2= (13)^2

For an answer of
(x+1)^2 + (y-1)^2 = 13

 

[tkhunny]

(x+1)^2 + (y-1)^2= (13)^2
(x+1)^2 + (y-1)^2 = 13

Hmmm...It looks like you got there, but I'm not quite sure how you did that.

I think you mean for the top one (x+1)^2 + (y-1)^2= (sqrt(13))^2

Right? Then we're good. Good work.