4a. Given two circles $k_{1}$ and $k_{2}$ that intersect at the different points $P$ and $Q$. Construct a line $A B$ running through $P$ with its endpoints at $k_{1}$ and $k_{2}$ so that the product $|A P| \cdot|P B|$ is maximal.

## Solution
Let $t_{1}$ and $t_{2}$ be the tangents to $k_{1}$ and $k_{2}$ in $P$. These divide the plane into four regions, one of which contains the point $Q$. Let $G$ be the union of the two opposite areas which do not contain the point $Q$. Obviously, the line $A B$ must run entirely in $G$ because otherwise $A=P$ or $B=P$ and the product $|A P| \cdot|P B|$ disappears. Let $\alpha$ be the opening angle of $G$ and let $0 \leq \varphi \leq \alpha$ be the angle between $A B$ and $t_{1}$. Let $r_{1}, r_{2}$ be the radii of the circles $k_{1}, k_{2}$. According to the tangent-center angle theorem, the chord $A P$ in $k_{1}$ has the center angle $2 \varphi$ and therefore has the length $|A P|=2 r_{1} \sin (\varphi)$. The angle between $P B$ and $t_{2}$ is equal to $\alpha-\varphi$ and, analogously, we obtain $|P B|=2 r_{2} \sin (\alpha-\varphi)$. To maximize the product $|A P| \cdot|P B|$, we need to find the maximum of $\sin (\varphi) \cdot \sin (\alpha-\varphi)$. According to the product-to-sum formula

$$
\sin (\varphi) \cdot \sin (\alpha-\varphi)=\frac{1}{2}(\cos (2 \varphi-\alpha)-\cos (\alpha))
$$

This is maximum when the first summand assumes the largest possible value, i.e. when $\varphi=\alpha / 2$. The distance you are looking for therefore lies on the bisector of the two tangents in $G$.