89. Points $A, B$ and $C$ are located on the same line ( $B$ - between $A$ and $C$ ). Find the geometric locus of points $M$ such that $\operatorname{ctg} \angle A M B+\operatorname{ctg} \angle B M C=k$.
89. Draw a perpendicular to $B M$ at point $M$; let $P$ be the point of intersection of this perpendicular and the perpendicular erected to the original line at point $B$. We will show that the magnitude $|P B|$ is constant. Let $\angle M B C=\varphi$; through $K$ and $L$ denote the feet of the perpendiculars dropped from $A$ and $C$ to $M B$. By the condition $\frac{|M K|}{|K A|}+\frac{|L M|}{|L C|}=k$,

![](https://cdn.mathpix.com/cropped/2024_05_21_35c23d544adcc44dca41g-132.jpg?height=385&width=323&top_left_y=1250&top_left_x=795)

Fig. 25 but $\quad|L C|=|B C| \sin \varphi, \quad|A K|=|B A| \sin \varphi$. Therefore, $\frac{|M K|}{|B A| \sin \varphi}+\frac{|L M|}{|B C| \sin \varphi}=k \Leftrightarrow \frac{|B M| \pm|B K|}{|B A| \sin \varphi}+\frac{|B M| \mp|B L|}{|B C| \sin \varphi}=$
$=k \Leftrightarrow \frac{|B M|}{\sin \varphi}\left(\frac{1}{|B A|}+\frac{1}{|B C|}\right) \pm\left(\frac{|B K|}{|B A| \sin \varphi}-\frac{|B L|}{|B C| \sin \varphi}\right)=k \Leftrightarrow \frac{|B M|}{\sin \varphi}=$
$=\frac{k|B A| \cdot|B C|}{|B A|+|B C|} \Leftrightarrow|P B|=\frac{k|B A| \cdot|B C|}{|B A|+|B C|}$, which is what we needed to show. Therefore, the required geometric locus of points is two circles tangent to the line $A C$ at point $B$, with diameters equal to $\frac{k|B A| \cdot|B C|}{|B A|+|B C|}$.