On the circle $O$, there are two points $M$ and $N$ which are symmetric with respect to a diameter, and a point $A$ lies on this diameter. $1^{0}$. If the point $A$ is fixed and the points $M$ and $N$ move on the circle $O$ in such a way that they remain symmetric with respect to the diameter $OA$, determine the geometric locus of the second intersection point $P$ of the line $OM$ and the circle $OAN$.

$2^{0}$. Determine the geometric locus of the second intersection point of two circles, one of which passes through $O$ and touches the circle $C$ at point $P$, and the other passes through $O$ and touches the circle $O$ at point $M$.
The equation of circle $O$ is

$$
x^{2}+y^{2}=r^{2}
$$

$A(c, 0), \quad M(\xi, \eta), \quad N(\xi,-\eta)$

Since $M$ and $N$ are on the circumference of circle $O$:

$$
\xi^{2}+\eta^{2}=r^{2}
$$

The equation of circle $AON$ is:

$$
x^{2}+y^{2}+2 \alpha x+2 \beta y=0
$$

because it passes through point $O$. It also passes through point $A$, so:

$$
2 \alpha=-c
$$

The common power line of the 1st and 2nd circles is:

$$
c x-2 \beta y-r^{2}=0,
$$

which intersects the diameter $AO$ at a constant point $A^{\prime}\left(\frac{r^{2}}{c}, 0\right)$. If the endpoints of diameter $AO$ are $G$ and $H$, then $\left(G A H A^{\prime}\right)=-1$, because:

$$
A G=(r-c), \quad H A=(r+c), \quad A^{\prime} G=\frac{r(r-c)}{c}, \quad A^{\prime} H=\frac{r(r+c)}{c}
$$

thus:

$$
(A G: A H):\left(A^{\prime} G: A^{\prime} H\right)=-1,
$$

which means: $A^{\prime}$ is the harmonic conjugate of $A$, and together with $A$ it changes its position in a known manner.

The common power line of the circles always passes through point $N$ as well, so:

$$
c \xi+2 \beta \eta-r^{2}=0
$$

The equation of diameter $OM$ is:

$$
\eta x-\xi y=0
$$

From 3) and 4):

$$
\xi=\frac{r^{2} x}{c x+2 \beta y}, \quad \eta=\frac{r^{2} y}{c x+2 \beta y}
$$

Substituting the found values of $\xi$ and $\eta$ into 1)a), and considering that from 2):

$$
2 \beta y=-\left(x^{2}+y^{2}-c x\right),
$$

thus

$$
c x+2 \beta y=-\left(x^{2}+y^{2}-2 c x\right),
$$

the equation of the desired geometric locus is:

$$
\left(x^{2}+y^{2}-2 c x\right)^{2}-r^{2}\left(x^{2}+y^{2}\right)=0
$$

The geometric locus is a quartic curve series, whose members vary in shape depending on the position of point $A$ (the value of $c$).

# Analysis of the general curve.

From equation 5), it can be seen that point $O$ is always a point of the curve, and it is either a node, an isolated double point, or a cusp, because for any value of $c$:

$$
y^{2}=0
$$

substituting:

$$
x^{2}\left[(x-2 c)^{2}-r^{2}\right]=0,
$$

thus $x^{2}=0$ always results. The abscissas of the other two intersection points with the diameter $OA$ are:

$$
\begin{aligned}
(x-2 c)^{2}-r^{2} & =0 \text {, from which } \\
x=2 c+r, \quad x & =2 c-r .
\end{aligned}
$$

The endpoints of the diameter perpendicular to $OA$ are also always points of the curve, because for $x=0$:

$$
y^{2}\left(y^{2}-r^{2}\right)=0,
$$

thus from the last factor:

$$
y= \pm r
$$

results.

The curve is always closed, symmetric with respect to the diameter $OA$, and depending on the position of point $A$:

$$
r>2 c>0
$$

if $O$ is an isolated double point.

The closer $O$ gets to $A$, the more the curve approaches the circle, and for $c=0$ it coincides with it.

For $r=2 c$, point $O$ is a cusp of the curve, and the diameter $OA$ is tangent to the curve at point $O$.

For $2 c>r$, the curve has a loop, and $O$ becomes a node. If $c=r$, i.e., point $A$ is the endpoint of the diameter $OA$, point $A$ is also a point of the curve. As $A$ moves away, the loop part becomes larger, and if $A$ goes to infinity: the curve, as the diameter perpendicular to $OA$, degenerates into a double line.

To determine the two tangents to the node, let the equation of the tangent be:

$$
y=m x
$$

which line has three points in common with the curve at $O$ if it is a tangent:

Substituting this into the equation of the curve:

$$
x^{2}\left[\left(\left(1+m^{2}\right) x-2 c\right)^{2}-r^{2}\left(1+m^{2}\right)\right]=0
$$

which equation must be satisfied by $x=0$ three times, for the determination of $m$:

$$
r^{2}\left(1+m^{2}\right)=4 c^{2}
$$

results, i.e.,

$$
m= \pm \frac{\sqrt{4 c^{2}-r^{2}}}{r}
$$

always, which equation also shows: when $O$ will be a singular point of the curve in one way or another.

In every other point, the tangent of the angle of the tangent is formed in a known way:

$$
\tan \alpha=\frac{2\left(x^{2}+y^{2}-2 c x\right)(x-c)-r^{2} x}{y\left(r^{2}-2\left(x^{2}+y^{2}-2 c x\right)\right)}
$$

or in terms of the coordinates of point $M$:

$$
\tan \alpha=\frac{2 c r^{2}+r^{2} \xi-4 c \xi^{2}}{\eta\left(4 c \xi-r^{2}\right)}
$$

## II.

To determine the geometric locus desired in the second part of the problem, first express the coordinates of $P(x, y)$ in terms of the coordinates of $M$ $(\tau, \mu)$, starting from the

$$
\tau=-\frac{r^{2} x}{\left(x^{2}+y^{2}-2 c x\right)}, \quad \mu=-\frac{r^{2} y}{\left(x^{2}+y^{2}-2 c x\right)}
$$

relationships and the equation of diameter $OM$

$$
\mu x-\tau y=0
$$

Considering that

$$
\tau^{2}+\mu^{2}=r^{2}
$$

always

$$
x=\frac{2 c \tau-r^{2}}{r^{2}} \tau, \quad y=\frac{2 c \tau-r^{2}}{r^{2}} \mu
$$

Given the already known value of the tangent of the angle of the tangent at point $P$, the equation of the circle of the radius of the tangent is

$$
y-\frac{2 c \tau-r^{2}}{r^{2}} \mu=\frac{\mu\left(r^{2}-4 c \tau\right)}{2 c r^{2}+r^{2} \tau-4 c \tau^{2}}\left(x-\frac{2 c \tau-r^{2}}{r^{2}} \tau\right)
$$

The equation of the perpendicular bisector of the segment $OP$ through the midpoint of diameter $OM$ is:

$$
y-\frac{2 c \tau-r^{2}}{2 r^{2}} \mu=-\frac{\tau}{\mu}\left(x-\frac{2 c \tau-r^{2}}{2 r^{2}} \tau\right)
$$

From 6) and 7), the coordinates of the center of the tangent circle are:

$$
p=\frac{2 c-\tau}{2}, \quad q=-\frac{\mu}{2}
$$

Based on these, the equation of the circle tangent at point $P$ and passing through $O$ (not the curvature or osculating circle) is:

$$
x^{2}+y^{2}-(2 c-\tau) x+\mu y=0
$$

The equation of the circle with diameter $\overline{OM}$ is:

$$
x^{2}+y^{2}-\tau x-\mu y=0
$$

By adding the two equations and eliminating the indeterminates $\tau$ and $\mu$, the equation of the desired geometric locus is:

$$
x^{2}+y^{2}-c x=0
$$

which is the circle with diameter $\overline{AC}$.