$\mathrm{P}(6,2 \sqrt{3})$ is a point on $\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1 .$ If $\mathrm{Q}$ is the projection of $\mathrm{P}$ on transverse axis, the point of contact of the tangent drawn from $\mathrm{Q}$ to the auxiliary circle of given hyperbola is
(a) $\left(\pm \frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)$
(b) $\left(\pm \frac{3}{2}, \pm \frac{3 \sqrt{3}}{2}\right)$
(c) $\left(\frac{3}{2}, \pm \frac{3 \sqrt{3}}{2}\right)$
(d) $\left(-\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)$