Given points $A, B, C$ on the plane and an angle $\varphi$. Construct a circle $k$ passing through points $A$ and $B$ such that $C$ sees the circle $k$ at an angle $\varphi$, that is, the angle between the tangents drawn from $C$ to $k$ is $\varphi$.
I. solution. It is sufficient to construct a figure similar to the desired one, as an appropriate similarity transformation will place it in the correct position. Draw a $\varphi$ angle domain with vertex $C$ and a circle $k'$ tangent to its sides. If we now construct the points $A'$ and $B'$ on the circumference of $k'$ such that the figures $C A' B'$ and $C A B$ are similar (1st figure), then the rotation-dilation centered at $C$ that maps $A'$ to $A$ will map $B'$ to $B$, and the circle $k'$ to a circle $k$ passing through $A$ and $B$, while the angle under which $k$ is seen from $C$ remains equal to $\varphi$. Thus, $k$ is a solution to the problem.

$1988-05-208-1 . \mathrm{eps}$

# 1.a figure 

$1988-05-208-2 . \mathrm{eps}$

## $1 . b$ figure

If the figures $C A' B'$ and $C A B$ are similar, then the point $A'$ is mapped to $B'$ by a rotation-dilation $\mathcal{F}$ centered at $C$ with angle $ACB$. (In figure 1/b, where $A, B$, and $C$ lie on a straight line, $\mathcal{F}$ is a central similarity.) The transformation $\mathcal{F}$ maps the circle $k'$ passing through $A'$ to a circle $k''$ passing through $B'$. Therefore, the point $B'$ is obtained as the intersection of the circles $k'$ and $k''$, if it exists. The point $A'$ is then obtained by the intersection of the circle $k'$ with the ray starting at $B'$ and forming the angle $CBA$ with $CB'$ (1.a figure).

The number of solutions is 0, 1, or 2, depending on how many common points the circles $k'$ and $k''$ have.

The two auxiliary circles, $k'$ and $k''$, intersect or touch each other if and only if the sum of their radii is at least as large as the distance between their centers. (The two circles cannot contain each other, as $C$, the center of similarity, is an external point to both.) Let the radii be $r'$ and $r''$, and the centers $O'$ and $O''$. Then

$$
\begin{aligned}
r' & =O' C \cdot \sin \frac{\varphi}{2} \\
r'' & =r' \cdot \frac{BC}{AC}
\end{aligned}
$$

From the similarity of the figures $O'CO''$ and $ACB$, we also have

$$
O' O''=O' C \cdot \frac{AB}{AC}
$$

As we said, a solution to the problem exists if and only if

$$
\begin{gathered}
r' + r'' \geqq O' O'', \quad \text { that is } \\
O' C \cdot \sin \frac{\varphi}{2}\left(1+\frac{BC}{AC}\right) \geqq O' C \cdot \frac{AB}{AC}
\end{gathered}
$$

After rearranging, we get the condition

$$
\sin \frac{\varphi}{2} \geqq \frac{AB}{AC + BC}
$$

$$
\begin{gathered}
\text { 1988-05-209-1.eps } \\
\text { 2.a figure }
\end{gathered}
$$

$1988-05-209-2 . \mathrm{eps}$

$2 . b$ figure

If equality holds in (1), then $k'$ and $k''$ touch each other, and the problem has one solution. If the left side is larger, then the circles intersect, and the number of solutions is two. In the case shown in figure 2/b, that is, when the points lie on a straight line and $C$ is not in the interval $AB$ - otherwise, of course, there is no solution - the two solutions are congruent.

II. solution. Let the center of the circle to be constructed be $O$, its radius $r$, and use the additional notations of figure 3.

In the right triangle $COE$,

$$
\sin \frac{\varphi}{2}=\frac{r}{CO}
$$

Since $EO=AO=r$, the condition for the center $O$ of the circle to be constructed is

$$
\frac{AO}{OC}=\sin \frac{\varphi}{2}
$$

It is known that the geometric locus of points in the plane whose distances from two given points - $A$ and $C$ - have a given ratio, which is a positive number different from 1, is a circle. (The proof can be found, for example, on page 124 of György Hajós's book "Introduction to Geometry"). The point $O$ is therefore on the circle of Apollonius corresponding to the points $A$ and $C$ with ratio $\sin \frac{\varphi}{2}$. And since $BO=r$ and thus

$$
\frac{BO}{OC}=\sin \frac{\varphi}{2}
$$

is also true, the point $O$ is also on the Apollonius circle corresponding to the points $B$ and $C$ with ratio $\sin \frac{\varphi}{2}$ (similarly to the perpendicular bisector of the segment $AB$).

## $1988-05-210-1$. eps

4. figure

By drawing two of the three lines - two circles and one line - we get the center of the circle to be constructed, and with the knowledge of one of its points, $A$ or $B$, the circle itself (figure 4). The problem has 2, 1, or 0 solutions depending on whether the two circles intersect, touch, or avoid each other. From this, with some calculation, (1) follows as the necessary and sufficient condition for solvability.