Given a circle $O$ and two points $A$ and $B$ in its plane. Connecting a point $M$ on the circle to points $A$ and $B$, the lines $AM$ and $BM$ intersect the circle again at points $C$ and $D$, respectively. Determine the point $M$ such that $CD \parallel AB$!
I. Solution. Let us assume that we have solved our problem: point $M$ meets the requirements of the problem, such that $C D \| A B$. Now let us consider the circumcircle $\gamma$ of $\triangle A M B$. Since $C D \| A B$, it is clear that $\frac{M A}{M C}=\frac{M B}{M D}$, meaning the two circles are in a similar position with respect to $M$; however, $M$ can only be a similarity center of the circles $O$ and $\gamma$ if it is the point of tangency of the two circles. Therefore, we need to construct a circle $\gamma$ that passes through points $A$ and $B$ and is tangent to circle $O$. There are two such circles, unless points $A$ and $B$ are on the same side of circle $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_af6dfd2dd88232815d45g-1.jpg?height=526&width=588&top_left_y=409&top_left_x=758)

Construct a circle $k$ that passes through points $A$ and $B$ and intersects circle $O$ at points $K$ and $L$. Then the line $K L$ is the radical axis of circles $k$ and $O$, intersecting $A B$ at point $P$, whose power with respect to circles $k$ and $O$ is equal, and also equal with respect to circles $\gamma$ and $k$. Therefore, the power of $P$ with respect to circles $\gamma$ and $O$ is also the same: thus, the radical axis of $\gamma$ and $O$, i.e., their common tangent, must pass through the fixed point $P$.

According to the above, we determine point $P$ and draw tangents from $P$ to $O$, with points of tangency $M_{1}$ and $M_{2}$: these will be the sought points $M$.

This construction cannot be performed if, for example, $A$ is outside the circle and $B$ is inside the circle. Indeed, if we connect any point $M$ on the circle to $A$ and $B$, points $C$ and $M$ are on the same side of line $A B$, but points $D$ and $M$ are on opposite sides of $A B$, and thus $C D$ will intersect $A B$.

József Klein (izr. g. VII. o. Debrecen).

II. Solution. If $M$ meets the requirements of the problem, such that $A B \| C D$, then

$$
\frac{A C}{A M}=\frac{B D}{B M} \quad \text { or } \quad \frac{A C \cdot A M}{\overline{A M}^{2}}=\frac{B D \cdot B M}{\overline{B M}^{2}}
$$

However, $\overline{A C} \cdot \overline{A M}=\alpha^{2}$ is the power of point $A$ with respect to circle $O$, and $\overline{B D} \cdot \overline{B M}=\beta^{2}$ is the power of point $B$ with respect to circle $O$; these can be considered known and should have the same sign. Therefore,

$$
\frac{\overline{B M}^{2}}{\overline{A M}^{2}}=\left(\frac{\alpha}{\beta}\right)^{2}, \quad \frac{B M}{A M}=\frac{\alpha}{\beta},
$$

meaning the ratio of the distances from point $M$ to points $A$ and $B$ is constant, and thus the geometric locus of point $M$ is the Apollonian circle corresponding to the value $\frac{\alpha}{\beta}$ for points $A$ and $B$, which intersects circle $O$ at points $M_{1}$ and $M_{2}$.[^0]


[^0]:    ${ }^{1}$ Or both are inside the circle, or both are outside the circle!