Let $C$ be a circle with the middle point $O$ and the radius $r$, we say that the points $P$ and $P'$ are inverse points with respect to $C$ if:
$1.$ $|OP|·|OP'|=r^2$
$2.$ $P$ and $P'$ are on the same line that starts in $O$
Assume that $P$ is located outside $C$ and draw the circle $C'$ with the diameter $\mid OP \mid$. $A$ is one of the intersection points between the circles $C$ and $C'$.Draw the normal line $N$ from $A$ towards $\mid OP \mid$ and let $P'$ be the point where $N$ intersects with $\mid OP \mid$ (see figure below). Prove that $P$ and $P'$ are inverse points with respect to $C$.
I drew $ΔOAP$ and $OAP'$ and proved that they are similar since they have 3 congruent angels and then I used that $\frac{|OA|}{|OP|}=\frac{|OP'|}{|OA|}$, where $|OA|$ is the radius in circle $C$ and then I got that $|OP|·|OP'|=r^2$, which is correct but that wasn't the proof that we were supposed to use so not really sure how to prove it in a different way.
1 Answer
Let B be the second point of intersection of $C$ and $C'$. Do an inversion in $C$. Points A and B are fixed under that inversion. The image of the circle $C'$ under that inversion is a line, since $C'$ passes through the center of inversion. That line contains points $A$ and $B$, so the image of $C'$ is the line $AB$ and the image of $P$ is $P'$
