4. Determine the geometric locus of the centers of the circles that pass through the point $A(3,0)$ and are tangent to the circle $x^{2}+y^{2}=25$.

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Solution. Let the point $M(p, q)$ belong to the desired geometric locus of points from the problem. Then there exists a circle with center at $M$

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

which passes through the point $A$ and is tangent to the given circle.

Since the circle (1) passes through the point $A$, we get

$$
(3-p)^{2}+(0-q)^{2}=r^{2}
$$

i.e.

$$
r^{2}-p^{2}-q^{2}=9-6 p
$$

Since the circle (1) is tangent to the given circle, the system

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=25 \\
x^{2}+y^{2}-2(p x+q y)=r^{2}-p^{2}-q^{2}
\end{array}\right.
$$

which, considering (2), can be written in the form

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=25 \\
x^{2}+y^{2}-2(p x+q y)=9-6 p
\end{array}\right.
$$

has a unique solution. If we subtract the second equation of system (3) from the first equation, we get $q y=8+3 p-p x$.

The unique point of the desired geometric locus of points for which $q=0$ is the point $M_{\circ}(-1,0)$. We will consider the desired locus of points without this point, i.e., we will assume that $q \neq 0$. Then $y=\frac{8+3 p-p x}{q}$.

If we substitute this value into the first equation of system (3), we get a quadratic equation in $x$

$$
x^{2}-\frac{2 p(8+3 p)}{p^{2}+q^{2}} x+\frac{(8+3 p)^{2}-25 q^{2}}{p^{2}+q^{2}}=0
$$

Since system (3) has a unique solution, and equation (4) has a unique solution, the discriminant of the quadratic equation must be equal to zero, i.e.

$$
4 \frac{p^{2}(8+3 p)^{2}}{\left(p^{2}+q^{2}\right)^{2}}-4 \frac{(8+3 p)^{2}-25 q^{2}}{p^{2}+q^{2}}=0
$$

This equality transforms into $16 p^{2}+25 q^{2}-48 p-64=0$ or

$$
\frac{\left(p-\frac{3}{2}\right)^{2}}{\left(\frac{5}{2}\right)^{2}}+\frac{q^{2}}{4}=1
$$

The point $M_{\circ}(-1,0)$ also satisfies this equation. Therefore, the desired geometric locus of points is an ellipse.