10,11

Spheres with centers at points $O_1$ and $O_2$ and radii 3 and 1, respectively, touch each other. Through point $M$, which is 3 units away from $O_2$, two lines are drawn, each tangent to both spheres, with the points of tangency lying on the lines on the same side of point $M$. Find the angle between the tangents, given that one of them forms a $45^{\circ}$ angle with the line $O_1 O_2$.
Let a line forming an angle of $45^{\circ}$ with the line $O 1 O 2$ touch the given spheres at points $B$ and $A$ respectively (Fig.1). Consider the tetrahedron $O 1 O 2 A B$. Extend it to a parallelepiped $A C B D A 1 O 1 B 1 O 2$ (Fig.2) by drawing pairs of parallel planes through the opposite edges of the tetrahedron (edges $A A 1, C O 1, B B 1, D O 2$ are parallel). Then $A B \perp A O 2$ and $A B \perp O 2 B$, so the line $A B$ is perpendicular to the plane $A A 1 O 2 D$ ($A B$ is perpendicular to two intersecting lines $A O 2$ and $D A 1$ of this plane). Therefore, $A B \perp A D$. Let $N$ be the point of intersection of the diagonals $A B$ and $C D$ of the parallelogram $A C B D$. These diagonals intersect at an angle of $45^{\circ}$ ($C D$ is parallel to $O 1 O 2$), and since $\angle D A N=90^{\circ}$ and $D N=N C=2$, then $A B=2 A N=2 \sqrt{2}$. If we now replace points $A$ and $B$ with the points of tangency of the second given tangent to the spheres, then by reasoning similarly, we can prove that the second tangent also forms an angle of $45^{\circ}$ with the line $O 1 O 2$. From the right triangle $M O 2 A$ we find that

$$
M A=\sqrt{M O_{2}^{2}-A O_{2}^{2}}=\sqrt{9-1}=2 \sqrt{2}
$$

Draw a plane through the given tangents. It will intersect the spheres along non-intersecting circles, one of which is outside the other, and the given tangents will be common external tangents drawn to these circles from point $M$. Let $O 1^{\prime}$ and $O 2^{\prime}$ be the orthogonal projections of points $O 1$ and $O 2$ onto the drawn plane. Then $O 1^{\prime}$ and $O 2^{\prime}$ are the centers of the circles obtained in the section. Denote the radius of the smaller circle $O 2^{\prime} A=t$. Since $A B=A M=2 \sqrt{2}$, the radius of the second circle $O 1^{\prime} B=2 t$. Let $O 2, O 2=x, 01$ 'O1 $=y$ (Fig.1). Then

$$
x 2=1-t 2, y 2=9-4 t 2 \text {. }
$$

Suppose points $O 1$ and $O 2$ are on the same side of the cutting plane (Fig.3). From the right trapezoid O1'O2'O2O1 (Fig.4) we find that

$$
O 1^{\prime} O 2{ }^{\prime} 2=16-(y-x) 2=16-\left(\sqrt{9-4 t^{2}}-\sqrt{1-t^{2}}\right) 2=6+5 t 2+2 \sqrt{4 t^{1}-13 t^{2}+9}
$$

From the right trapezoid $O 1^{\prime} O 2^{\prime} A B$ we find that

$$
O 1^{\prime} O 2^{\prime}=t 2+8
$$

Solve the equation

$$
6+5 t 2+2 \sqrt{4 t^{1}-13 t^{2}+9}=t 2+8
$$

or

$$
\sqrt{4 t^{1}-13 t^{2}+9}=1-2 t 2
$$

After squaring and simplifying, we get an equation from which we find that $t 2=\frac{8}{9}$. This root does not satisfy the original equation. Therefore, points $O 1$ and $O 2$ cannot lie on the same side of the cutting plane. If, however, points $O 1$ and $O 2$ lie on opposite sides of the cutting plane, then reasoning similarly, we find that $t=\frac{2 \sqrt{2}}{3}$, from which $\operatorname{tg} \frac{\varphi}{2}=\frac{\mathbf{5}}{2 \sqrt{2}}=\frac{1}{3}$, where $\phi-$ is the desired angle.