I'm really stuck.
I'm learning about circle inversion. More specifically, I was trying to understand how to derive the inversion formula for a circle, which seems to be explained here.
http://classes.yale.edu/fractals/circinvfrac/InvFormulas/InvFormulas2Mv.html#CircAnchor
How is it that the unit vector is $\frac{1}{\sqrt{(a-c)^2+(b-d)^2}}$?
Why is it not just $\sqrt{(a-c)^2+(b-d)^2}(a-c,b-d)$ itself?
It may sound like a really stupid question, but I've done quite a bit of Googling on "unit vectors", and nothing seemed to really explain this calculation.
Thanks so much!
$\frac{1}{\sqrt{(a-c)^2+(b-d)^2}}$ is the length of the vector between the two circle centers.
The vector itself is $(a-c,b-d)$. The unit vector, then, is the original vector divided by its length:
$$\frac{1}{\sqrt{(a-c)^2+(b-d)^2}}(a-c,b-d).$$