A tangent to the ellipse $\frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{4}=1$ is met by the tangents at the end of the major axis at the points P and Q. If the circle on PQ as diameter passes through fixed point R, then R may be:

A tangent to ellipse is $\frac{\mathrm{x}}{3}\cos \mathrm{\varphi }+\frac{\mathrm{y}}{2}\sin \mathrm{\varphi }=1$

It intersects x = 3 and x = -3 at points

$\mathrm{P}\left(3,\frac{2(1-\cos \mathrm{\varphi })}{\sin \mathrm{\varphi }}\right)\mathrm{\&}\mathrm{Q}\left(-3,\frac{2(1+\cos \mathrm{\varphi })}{\sin \mathrm{\varphi }}\right)$

The circle with PQ as diameter is 

$\begin{array}{l}(\mathrm{x}+3)(\mathrm{x}-3)+\left(\mathrm{y}-\frac{2(1-\cos \mathrm{\varphi })}{\sin \mathrm{\varphi }}\right)\left(\mathrm{y}-\frac{2(1+\cos \mathrm{\varphi })}{\sin \mathrm{\varphi }}\right)=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-5-\left(\frac{4}{\sin \mathrm{\varphi }}\right)\mathrm{y}=0\end{array}$

It is a family of circles passing through the intersection of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-5=0$ and the line y = 0, which is $(\pm \sqrt{5},0)$