AC is transverse common tangent to two circles with centres P and Q and radius 6 cm and 3 cm at the point A and C respectively. If AC cuts PQ at the point B and AB = 8cm then the length of PQ is :

In $\Delta \mathrm{APB}$ and $\Delta \mathrm{BCQ}$,

$\angle \mathrm{PAB}=\angle \mathrm{BCQ}=90^{\circ}$

$\angle \mathrm{PBA}=\angle \mathrm{QBC}$

By AA - similarity,

$\triangle \mathrm{APB} \sim \Delta \mathrm{BQC}$

$\therefore \quad \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AP}}{\mathrm{QC}}$

$\Rightarrow \frac{8}{\mathrm{BC}}=\frac{6}{3}$

$\Rightarrow \mathrm{BC}=\frac{8 \times 3}{6}=4 \mathrm{~cm} .$

$\therefore \quad \mathrm{PQ}=\sqrt{\mathrm{AC}^{2}+\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)^{2}}$

$=\sqrt{(8+4)^{2}+(6+3)^{2}}=\sqrt{12^{2}+9^{2}}=\sqrt{144+81}$

$=\sqrt{225}=15 \mathrm{~cm} .$