The point $P$ is located outside the circle with center $O$. The lines $l_1$ and $l_2$ pass through the point $P$. The line $l_1$ touches the circle in the point $A$ and the line $l_2$ intersects the circle at the points $B$ and $C$. The tangents of the circle that pass through point $B$ and $C$ intersect at the point $X$. How can I prove that $PO$ is perpendicular to $AX$?


I think it is possible to prove this by drawing the perpendiculars $AA'$ and $XX'$. All that remains is to prove that the points $A'$ and $X'$ coincide. The only question is: how do I do that?enter image description here

  • $\begingroup$ Try to prove that if it is perpendicular than the point P exist. $\endgroup$ – Moti Feb 19 '18 at 17:01

This answer uses coordinate geometry.

We may suppose that the equation of the circle is $x^2+y^2=1$ with $P(p,0)$ where $p\lt -1$.

Let $A(a_x,a_y)$. Then, we have $$a_x\cdot p+a_y\cdot 0=1\implies a_x=\frac 1p$$

If the equation of the line $BC$ is $y=m(x-p)$, then we have $$B(x_b,m(x_b-p)),\quad C(x_c,m(x_c-p))$$

So, since $X$ is the intersection point of $$x_bx+m(x_b-p)y=1\quad\text{with}\quad x_cx+m(x_c-p)y=1$$ we get that the $x$-coordinate of $X$ is $\frac 1p$.

It follows that $PO$ is perpendicular to $AX$.


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