Solution:
Suppose that $P$ and $Q$ lie on the opposite sides of line joining $O_{1}$ and $O_{2}$. By symmetry we may assume that the configuration is as shown in the figure below. Then we have $K P > K O_{1} > K Q$ since $K O_{1}$ is the hypotenuse of triangle $K Q O_{1}$. This is a contradiction to the given assumption, and therefore $P$ and $Q$ lie on the same side of the line joining $O_{1}$ and $O_{2}$.

Since $K P = K Q$ it follows that $K$ lies on the radical axis of the given circles, which is the common tangent at $R$. Therefore $K P = K Q = K R$ and hence $K$ is the circumcenter of $\triangle P Q R$.

On the other hand, $\triangle K Q O_{1}$ and $\triangle K R O_{1}$ are both right-angled triangles with $K Q = K R$ and $Q O_{1} = R O_{1}$, and hence the two triangles are congruent. Therefore $\widehat{Q K O_{1}} = \widehat{R K O_{1}}$, so $K O_{1}$, and hence $P K$ is perpendicular to $Q R$. Similarly, $Q K$ is perpendicular to $P R$, so it follows that $K$ is the orthocenter of $\triangle P Q R$. Hence we have that $\triangle P Q R$ is equilateral.

Solution:
We again rule out the possibility that $P$ and $Q$ are on the opposite side of the line joining $O_{1} O_{2}$, and assume that they are on the same side.
Observe that $\triangle K P O_{2}$ is congruent to $\triangle K Q O_{1}$ (since $K P = K Q$). Therefore $O_{1} P = O_{2} Q = r$ (say). In $\triangle O_{1} O_{2} Q$, we have $\widehat{O_{1} Q O_{2}} = \pi / 2$ and $R$ is the midpoint of the hypotenuse, so $R Q = R O_{1} = r$. Therefore $\triangle O_{1} R Q$ is equilateral, so $\widehat{Q R O_{1}} = \pi / 3$. Similarly, $P R = r$ and $\widehat{P R O_{2}} = \pi / 3$, hence $\widehat{P R Q} = \pi / 3$. Since $P R = Q R$ it follows that $\triangle P Q R$ is equilateral.