Let the tangent of the circle with diameter $AB$ be $AT$. Determine the point $M$ on the circle for which the sum of the distances from the lines $AB$ and $AT$ is $l$, a given value. Investigate for which values of $l$ there is a solution and how many there are.
I. Solution. Let the distance of point $M$ from $AB$ be $MP=y$, and from $AT$ be $MQ=AP=x$. Clearly, $y=\sqrt{x(2R-x)}$, if $AB=2R$.

We need to solve the system of equations

$$
\begin{gathered}
x+y=l \ldots \\
y^{2}=x(2R-x) \ldots
\end{gathered}
$$

From (1), $y=l-x$; substituting this into (2). After appropriate rearrangement, we get

$$
f(x) \equiv 2x^{2}-2(l+R)x+l^{2}=0 \ldots
$$

The root of this equation satisfies the problem if it is real, positive, and less than $l$, and not greater than $2R$. $(l>0$ ! $)$

The roots of (3) are real if

$$
(l+R)^{2}-2l^{2} \geq 0, \quad (l+R)>l\sqrt{2} \quad \text{or} \quad R \geq l(\sqrt{2}-1).
$$

Multiplying both sides of the latter inequality by $(\sqrt{2}+1)$, we get

$$
l \leq R(\sqrt{2}+1) \ldots
$$

Now, let's examine the position of the roots of (3) relative to $0$, $l$, and $2R$ under the condition (4). We can note that if the roots are real, both are positive.

$$
f(0)=l^{2}>0, \quad f(2R)=(l-2R)^{2}>0, \quad f(l)=-l(2R-l)
$$

$1^{\circ}$. If $l<2R$, but $J$ is between $B$ and $L$, then the line $IJ$ intersects the circle at two points above the $X$-axis.

If $l=AL=R(\sqrt{2}+1)$, then the line $IJ$ is tangent to the circle at point $M'$. (The abscissa of $M'$ is: $R+\frac{l-R}{2}=\frac{l+R}{2}$). If $l>AL$, then a point $M$ satisfying the conditions does not exist: the line does not intersect the circle.