# 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.

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.

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.

• +1 Thank you. I'm not familiar with this property so will investigate. I wonder if there is another way since all I have learned about inversion so far is equivalent definitions of inverse points and the reciprocal theorem. Commented Aug 31, 2021 at 11:23

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.

• Thank you for your answer. Looking at the angles quickly becomes complicated. I will try that but I think the intended solution must use inversion as the problem comes from part of a book that is about inversion. Commented Aug 31, 2021 at 10:58
• Yeah that's why I haven't finished the calculation yet, as it's getting complicated soon. If this book is about inversion then probably inversion should be used for solving :D I'm not firm with inversion so for that I can't really help you Commented Aug 31, 2021 at 11:00