(a)OR(b)

In Figure $\displaystyle 3$, two circles with centres at O and O' of radii 2r and r respectively, touch each other internally at A. A chord AB of the bigger circle meets the smaller circle at C. Show that C bisects AB.

Figure $\displaystyle 3$

\[\angle O C A=90^{\circ} \] In \(\displaystyle \triangle s O C A\) and \(\displaystyle O C B\), we have \[O A=O B, \angle O C A=\angle O C B=90^{\circ} \] and \(\displaystyle O C=O C\) So, \(\displaystyle \triangle O C A \cong \triangle O C B\) \[\Rightarrow A C=B C \Rightarrow C \text { bisects } A B \] Or (b) In Figure $\displaystyle 4$, O is centre of a circle of radius $\displaystyle 5$ cm. PA and BC are tangents to the circle at \(\displaystyle A\) and \(\displaystyle B\) respectively. If \(\displaystyle O P=13 \mathrm{~cm}\), then find the length of tangents PA and BC. Figure $\displaystyle 4$ \[\begin{aligned} & P A^{2}=O P^{2}-O A^{2}=169-25 \\ & \Rightarrow P A=12 \mathrm{~cm} \end{aligned} \] Let \(\displaystyle B C=x \Rightarrow A C=x\) \[\therefore P C=12-x \] \[\begin{aligned} & O P \perp B C \Rightarrow(12-x)^{2}=x^{2}+8^{2} \\ & \Rightarrow x=\frac{10}{3} \mathrm{~cm} \\ & \therefore \mathrm{BC}=\frac{10}{3} \mathrm{~cm} \end{aligned} \]