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Mathematics · 2022 · 4 marks
CBSE 2022 · Region 3 · Set 3 · Q12
(a) In Figure 4, two circles with centres at O and O' of radii 2 r and r respectively, touch each other internally at A. A chord AB of the bigger…
In Figure $\displaystyle 4$, two circles with centres at O and O' of radii $\displaystyle 2$ r 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 $\displaystyle 4$
In Figure $\displaystyle 5$, O is centre of a circle of radius $\displaystyle 5$ cm . PA and BC are tangents to the circle at A and B respectively. If $\displaystyle \mathrm{OP}=13 \mathrm{~cm}$, then find the length of tangents PA and BC.
Figure $\displaystyle 5$
Marking-scheme solution
\[\angle O C A=90^{\circ}
\]
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
\[\begin{aligned}
& P A^{2}=O P^{2}-O A^{2}=169-25 \\
& \Rightarrow P A=12 \mathrm{~cm}
\end{aligned}
\]
Let \(\displaystyle B C=x \Rightarrow A C=x\)
\[\therefore P C=12-x
\]
\[\begin{aligned}
& O P \perp B C \Rightarrow(12-x)^{2}=x^{2}+8^{2} \\
& \Rightarrow x=\frac{10}{3} \mathrm{~cm} \\
& \therefore B C=\frac{10}{3} \mathrm{~cm}
\end{aligned}
\]
\[\begin{aligned}
& P A^{2}=O P^{2}-O A^{2}=169-25 \\
& \Rightarrow P A=12 \mathrm{~cm}
\end{aligned}
\]
Let \(\displaystyle B C=x \Rightarrow A C=x\)
\[\therefore P C=12-x
\]
\[\begin{aligned}
& O P \perp B C \Rightarrow(12-x)^{2}=x^{2}+8^{2} \\
& \Rightarrow x=\frac{10}{3} \mathrm{~cm} \\
& \therefore B C=\frac{10}{3} \mathrm{~cm}
\end{aligned}
\]More from Circles
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CBSE Class 10 Mathematics past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.