PQ and PR are two tangents to a circle with centre O and radius $\displaystyle 5$ cm. AB is another tangent to the circle at C which lies on OP. If $\displaystyle \mathrm{OP}=13 \mathrm{~cm}$, then find the length AB and PA.

Sol. \(\displaystyle \mathrm{OP}=13 \mathrm{~cm}, \mathrm{OQ}=5 \mathrm{~cm}\) \[\therefore P Q=\sqrt{169-25}=12 \mathrm{~cm} \] Let \(\displaystyle \mathrm{AC}=\mathrm{x}=\mathrm{AQ}\) \[\mathrm{PC}=13-5=8 \mathrm{~cm} \text { and } \mathrm{PA}=12-\mathrm{x} \] \(\displaystyle \mathrm{AC} \perp \mathrm{OP} \therefore(12-\mathrm{x})^{2}=\mathrm{x}^{2}+8^{2}\) \(\displaystyle \Rightarrow \mathrm{x}=\frac{10}{3}\) \(\displaystyle \mathrm{AB}=2 \mathrm{AC}=\frac{20}{3} \mathrm{~cm}\) or $\displaystyle 6.6$ cm (approx.) \(\displaystyle \mathrm{PA}=12-\frac{10}{3}=\frac{26}{3} \mathrm{~cm}\) or $\displaystyle 8.6$ cm (approx.)