Prove that the tangents at $X$ and $Y$ meet on the line $AB$ - circle inversion 
Let $P$ be a point outside a circle and let the tangents from $P$ touch the circle at $A$ and $B$. A line through $P$ intersects the circle in points $X$ and $Y$. Prove that the tangents at $X$ and $Y$ meet on the line $AB$.

I have drawn a diagram and I know that the intersection of $OP$ and $AB$ (call it $P_1$) is the inverse of $P$. Hence $OP\cdot OP_1=r^2=OX^2=OY^2=OA^2=OB^2$. If we let $T$ be the point of intersection of the tangents through $X$ and $Y$ then $XT=YT$. I have tried supposing that $T$ lies on $AB$ to see what I can deduce and using Pythagoras and the definition of inverse points a few times to get $XT=QT$. But I can't see how this might help.
I would be grateful for a hint.
 A: 
We use the property of radical axis. Draw a circle on P , AB is radical axis of two circles. the points of tangents on both circles are co-linear with the center of other circle.That is line XY passes the center P and line TN passes O. So the intersection of tangents on points X and Y is on AB ,also intersection of tangents on T and N is on AB.
A: Here's one way to solve it using inversion.

Call $O$ the center of the initial circle, and let $C$ be the intersection of the tangents at $X$ and $Y$.
Draw the circle $\omega$ through $O$, $A$ and $P$. Here are some immediate properties:

*

*$\omega$ is the image of $(AB)$ under inversion (since $B\in\omega$ by symmetry);

*any point $Z\in\omega$ satisfies $(OZ)\perp (ZB)$ (since, by symmetry, $[OP]$ is a diameter of $\omega$).

Since $(OC)\perp(XY)$, the intersection $Z$ of $(OC)$ and $(XY)$ thus lies on $\omega$.
By construction, $Z$ is the image of $C$ under inversion. $Z\in\omega$ is equivalent to saying that $C\in (AB)$, as desired.
A: I would go over angles. It's a bit to calculate and I haven't finished calculation yet, but it seems promising. Let $M$ be the mid of the circle and $F$ the intersection of the tangent at $X$ and $AB$. The question now is: Is $\measuredangle FYM = 90°$? That would mean that the tangent in $Y$ also intersects at F, which we want to show.
You can use a lot of useful angle calculations:

*

*The triangles $\triangle MYA, \triangle MYB, \triangle MXA, \triangle MXB$ are all isosceles

*The angle at the center is double the angle at the top, e.g. $\measuredangle YMA = 2 \cdot \measuredangle YBA$ (if $M$ is in the inside of the triangle $\triangle YBA$)

*$\measuredangle FXM = 90°$ as this is a tangent.

*Sum of angles inside a triangle is 180° and inside a foursided convex polygon it is 360°

*You can also use common knowledge about the behaviour of angles at intersecting lines

I hope these tips help to solve the problem. At least for me it looks pretty promising.
