I don't understand the concept of circle inversion.
$OP \cdot OP' = k^2$
For example, in a circle $x^2+y^2=k^2$. If I set a general point $P(x,y)$, why is its image $P'(\frac{xk^2}{x^2+y^2}, \frac{yk^2}{x^2+y^2})$?
Also, why does a line become a circle through O?
Sorry for my English, I'm not a native English speaker.
Note that $O, P, P'$ must be collinear. Therefore if we take $P = (x,y)$ we must have $P' = (ax,ay)$ for some positive integer $a$. Now from the condition that $OP \cdot OP' = k^2$ we must have:
$$k^2 = \sqrt{x^2 + y^2} \sqrt{a^2x^2 + a^2y^2} = a(x^2+y^2) \implies a = \frac{k^2}{x^2+y^2}$$
For the second part note that the transformation is an inverse of itself, so $x=\frac{x'}{(kx')^2+(ky')^2} $ and $y=\frac{y'}{(kx')^2+(ky')^2}$.
Now a line is given by $Ax + By + C= 0$. Substitute from above and you will get an equation of a circle.
$$0 = A\frac{x'}{(kx')^2+(ky')^2} + B\frac{y'}{(kx')^2+(ky')^2} + C$$
$$-(Ck)(x'^2+y'^2) = Ax' + By'$$
$$Ax' + By' + (Ck)(x'^2+y'^2) = 0$$
This is an equation of circle passing through the origin. Note that if $C=0$ then the line passes through the origin and is sent to itself.