Question

In the given figure line AB is tangent to both the circles touching at A and B. OA = 29, OB = 18 and OP = 61 then find AB.

Answer 1

Construction: Draw a perpendicular from centre P to radius OA meeting at M.

$$\begin{gathered} \angle \mathrm{OAB}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \angle \mathrm{PBA}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \mathrm{In}\mathrm{quadrialteral}\mathrm{ABPM},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{OAB}+\angle \mathrm{PBA}+\angle \mathrm{PMA}+\angle \mathrm{MPB}=360°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow \angle \mathrm{MPB}=360°-\angle \mathrm{OAB}-\angle \mathrm{PBA}-\angle \mathrm{PMA} \\ \Rightarrow \angle \mathrm{MPB}=360°-90°-90-90°=90° \end{gathered}$$

Since all angles of $\square$ABPM are 90$°$, it is rectangle.

$$\begin{gathered} \mathrm{AM}=\mathrm{BP}=18\mathrm{units}(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{are}\mathrm{equal}) \\ \mathrm{OM}=\mathrm{AO}-\mathrm{AM}=29-18=11\mathrm{units} \end{gathered}$$

$$\begin{gathered} \mathrm{So},\mathrm{in}\mathrm{right}∆\mathrm{OMP},\mathrm{we}\mathrm{have}: \\ \mathrm{PM}=\sqrt{{\mathrm{OP}}^2-{\mathrm{OM}}^2} \\ =\sqrt{{61}^2-{11}^2} \\ =\sqrt{3721-121} \\ =\sqrt{3600} \\ =60\mathrm{units} \end{gathered}$$

$\therefore \mathrm{PM}=\mathrm{AB}=60\mathrm{units}(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{are}\mathrm{equal})$