# Circle Inversion

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.

## 2 Answers

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.

• $k^2 \sqrt{x^2 + y^2} \sqrt{a^2x^2 + a^2y^2} = a(x^2+y^2)$ Why are they equal? – user506851 Nov 26 '17 at 18:51
• I missed the "=" sign in between. I've edited it now – Stefan4024 Nov 26 '17 at 18:52
• @MayaFarber You can check my update on the second part of the problem – Stefan4024 Nov 26 '17 at 18:59
• For $C=0$ a radial straight line segment through center is sent to another part of same radial straight line on either side of the circle $\Gamma$. – Narasimham Nov 26 '17 at 19:45

the main relation $OP \cdot OP' = k^2$ has a clear and nice geometrical meaning by means of Euclid theorem

• It's one of the most important property of inversion, but I don't think it adresses the OP's question. – Stefan4024 Nov 26 '17 at 19:00
• It adresses the question because by the construction it easy to prove that a "line become a circle through O" and all the other properties – user Nov 26 '17 at 19:04
• I'm really curios how can you prove it that easy. – Stefan4024 Nov 26 '17 at 19:07
• I mean that it's easy to understand why the circle passes through O – user Nov 26 '17 at 19:12
• That part is easy, but why it has to be a circle? – Stefan4024 Nov 26 '17 at 19:13