4. Find the smallest value of the parameter $a$, for which the system of equations

$$
\left\{\begin{array}{c}
\sqrt{(x-6)^{2}+(y-13)^{2}}+\sqrt{(x-18)^{2}+(y-4)^{2}}=15 \\
(x-2 a)^{2}+(y-4 a)^{2}=\frac{1}{4}
\end{array}\right.
$$

has a unique solution

#
# Solution.

The first equation of the system defines the locus of points $M(x, y)$ on the plane for which the sum of the distances from these points to points $A(6,13)$ and $B(18,4)$ is 15. Note that

$$
|A B|=\sqrt{12^{2}+9^{2}}=15
$$

Therefore, according to the triangle inequality, the locus of such points $M(x, y)$ is the segment $A B$.

The second equation is the equation of a circle with center at point $S(2 a, 4 a)$ and radius $\frac{1}{2}$.

A unique solution to the system is possible if and only if the circle intersects the segment $A B$ at exactly one point.

Obviously, a guaranteed unique point of intersection will occur in the case of the circle being tangent to the segment. This will happen when the distance from the point $S(2 a, 4 a)$ to the line containing the segment $A B$ is equal to the radius of the circle, and the point of tangency will fall on the segment $A B$. The equation of the line containing $A B$, as it is not difficult to establish, is $3 x+4 y-70=0$. According to the formula for the distance from a point to a line (one of the solution options):

$$
\frac{|3 \cdot 2 a+4 \cdot 4 a-70|}{\sqrt{3^{2}+4^{2}}}=\frac{1}{2}
$$

From this, we will obtain two possible values of the parameter $a$:

![](https://cdn.mathpix.com/cropped/2024_05_06_6714d21cf0c92eec56b3g-2.jpg?height=888&width=850&top_left_y=1635&top_left_x=543)

The center of the circle lies on the line $y=2 x$. The point $M\left(\frac{70}{11}, \frac{140}{11}\right)$ of intersection of the lines $y=2 x$ and $3 x+4 y-70=0$ lies on the segment $A B$. The angle $\angle OMB$ is acute, so the point of tangency of the line $3 x+4 y-70=0$ and the circle, whose center lies below the segment $A B$, will definitely fall on the segment $A B$. This occurs when $a=\frac{135}{44}$. If the center $S$ of the circle lies above the segment $A B$ (this happens when $a=\frac{145}{44}$), then additional reasoning is required. The point of tangency is the projection of the point $S\left(\frac{145}{22}, \frac{145}{11}\right)$ onto the line containing the segment $A B$. $H$ will fall on the segment $A M$ if $M H \leq A M$. We have:

$$
M H=\sqrt{S M^{2}-S H^{2}}=\sqrt{\left(\frac{145}{22}-\frac{70}{11}\right)^{2}+\left(\frac{145}{11}-\frac{140}{11}\right)^{2}-\frac{1}{4}}=\frac{1}{11}
$$

$$
A M=\sqrt{\left(\frac{70}{11}-6\right)^{2}+\left(\frac{140}{11}-13\right)^{2}}=\frac{5}{11} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6714d21cf0c92eec56b3g-3.jpg?height=618&width=1670&top_left_y=734&top_left_x=158)

Therefore, $M H<A M$, and the point of tangency $H$ lies on the segment $A B$.

At the same time, since $A M<\frac{1}{2}$, a unique solution is possible when the circle intersects the segment $A B$, but the point $A$ falls inside the circle. This will happen from the moment the circle intersects the segment at point $A$ until the moment of re-intersection at the same point $A$ (excluding these moments).

Let's find such positions of the point $S(2 a, 4 a)$ for which the distance from it to the point $A$ is $\frac{1}{2}$. We have:

$$
(2 a-6)^{2}+(4 a-13)^{2}=\frac{1}{4}
$$

From this,

$$
\left[\begin{array}{l}
a=\frac{13}{4} \\
a=\frac{63}{20}
\end{array}\right.
$$

Thus, for $a \in\left(\frac{63}{20}, \frac{13}{4}\right)$, the point of intersection will be unique, as will the solution to the system of equations.

Answer: $135 / 44$.