Given a circle $k$ and a diameter line $e$ of it. Determine the set of centers of those circles that touch $e$, and the distance of their closest point to $k$ is equal to their radius.
Let's set up a Cartesian coordinate system such that the center of $k$ is the origin $O$, the $x$-axis coincides with $e$, and the unit is the radius of $k$. In this coordinate system, we will provide the equation of the set of centers that satisfy the conditions.

It is sufficient to deal with the points in the half-plane $y \geq 0$, because the set of corresponding points will also be symmetric with respect to the $x$-axis, since both $k$ and $e$ are. Let $l$ be a circle that satisfies the conditions, with radius $r$ and center at the point $L(x, y)$. We distinguish two cases.

i) The point $L$ is outside $k$ (i.e., $x^{2}+y^{2}>1$). In this case, the point on $l$ closest to $k$ is the intersection of the segment $O L$ and $k$ (Figure 1), so $O L=2 r+1$. Since $l$ is tangent to the $x$-axis, $r=y$. Applying the Pythagorean theorem to the right triangle with hypotenuse $O L$ as shown in the figure:

$$
x^{2}+y^{2}=(2 r+1)^{2}
$$

From this, using the fact that $r=y$, we get the equation

$$
9\left(y+\frac{2}{3}\right)^{2}-3 x^{2}=1
$$

$L$ is therefore on the hyperbola given by equation (1). If we take a point on this hyperbola - the part marked with a thicker line in Figure 1 - which is outside $k$ and in the half-plane $y \geq 0$, then for this point $L^{\prime}\left(x^{\prime}, y^{\prime}\right)$, it holds that $\sqrt{{x^{\prime 2}}^{2}+y^{\prime 2}}=2 y^{\prime}+1$, meaning that if we draw a circle with radius $y^{\prime}$ around $L^{\prime}$, it will be tangent to the $x$-axis, and its closest point to $k$ will be $\left(2 y^{\prime}+1\right)-1-y^{\prime}=y^{\prime}$ away from $k$, thus satisfying the conditions of the problem.

ii) If the point $L$ is inside $k$ (i.e., $x^{2}+y^{2}<1$), then the point on $l$ closest to $k$ is the intersection of the ray $O L$ and $l$ (Figure 2). Applying the Pythagorean theorem again:

$$
x^{2}+y^{2}=(1-2 r)^{2}
$$

It is still true that $r=y$, so we get

$$
9\left(y-\frac{2}{3}\right)^{2}-3 x^{2}=1
$$

$L$ is therefore on the hyperbola given by equation (2). Just as in the previous case, it can be shown that every point of the hyperbola that is inside $k$ and in the half-plane $y \geq 0$ - the part marked with a thicker line in Figure 2 - satisfies the conditions of the problem.

In both cases, therefore, the set of centers that satisfy the conditions consists of segments of hyperbolas and their reflections across the $x$-axis, as shown in Figure 3. Since the hyperbolas given by equations (1) and (2) are reflections of each other across the $x$-axis, the desired set is actually composed of one branch of each hyperbola.

Kiss Gergely (Fazekas M. Fóv. Gyak. Gimn., 11th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_0bbe04b73ce5651b1d8eg-1.jpg?height=523&width=1285&top_left_y=1733&top_left_x=409)
![](https://cdn.mathpix.com/cropped/2024_05_02_0bbe04b73ce5651b1d8eg-2.jpg?height=1040&width=1282&top_left_y=130&top_left_x=411)