1. Let \( a \) be a positive real number. Find the geometric place of the centers of the circles that touch the circle \( x^{2}+y^{2}-2 a x=0 \) and the \( y \)-axis.
Solution. Let $k_{1}:(x-X)^{2}+(y-Y)^{2}=r^{2}$ be a circle whose center $M(X, Y)$ belongs to the desired geometric locus. Since the $y$-axis is tangent to $k_{1}$, it follows that $X=r$ or $X=-r$.

1) If $X=r$, we get

$$
k_{1}: x^{2}-2 x X+y^{2}-2 y Y+Y^{2}=0
$$

Now, from the system

$$
\left\{\begin{array}{l}
x^{2}-2 x X+y^{2}-2 y Y+Y^{2}=0 \\
x^{2}+y^{2}-2 a x=0
\end{array}\right.
$$

we get

$$
-2 x X-2 y Y+2 a x+Y^{2}=0
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_f3fdf1bc245e57626c19g-09.jpg?height=599&width=440&top_left_y=842&top_left_x=1025)

From this, if $Y \neq 0$, we have

$$
y=\frac{1}{2 Y}\left(-2 x X+2 a x+Y^{2}\right)=\frac{a-X}{Y} x+\frac{Y}{2}
$$

By substituting into the second equation of the system, we get

$$
x^{2}+\frac{(a-X)^{2}}{Y^{2}} x^{2}+(a-X) x+\frac{Y^{2}}{4}-2 a x=0 .
$$

Since the two circles have a unique intersection point, it follows that the discriminant of the last equation must be zero. Thus,

$$
(X+a)^{2}-Y^{2}\left(1+\frac{(a-X)^{2}}{Y^{2}}\right)=0
$$

By simplifying the last equation, we get $Y^{2}=4 a X$. Therefore, the desired geometric locus in this case is the parabola $y^{2}=4 a x$.

If $Y=0$, we get

$$
\left\{\begin{array}{l}
x^{2}-2 x X+y^{2}=0 \\
x^{2}+y^{2}-2 a x=0
\end{array}\right.
$$

From this, $2(a-X) x=0$. Since $a \neq X$ (otherwise the circles would coincide), we have $x=0$ and $y=0$. Therefore, the two circles are tangent. In this case, the desired geometric locus of points is the positive part of the $x$-axis.

2) If $X=-r$ and $Y=0$, working similarly as in 1) we get the system

$$
\left\{\begin{array}{l}
x^{2}+2 x r+y^{2}-2 y Y+Y^{2}=0 \\
x^{2}+y^{2}-2 a x=0
\end{array}\right.
$$

and then the equation

$$
\left(1+\left(\frac{a+r}{Y}\right)^{2}\right) x^{2}+(r-a) x+\frac{Y^{2}}{4}=0
$$

whose discriminant is $-Y^{2}-4 a r$ and is negative. Therefore, the circles have no intersection points. If $Y=0$, we get the negative part of the $x$-axis.

Finally, the desired geometric locus of points is the $x$-axis and the parabola $y^{2}=4 a x$.