CBSE 2025 · Region 2 · Set 1 · Q26 · 3 marks
Prove that the parallelogram circumscribing a circle is a rhombus.Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Prove that the parallelogram circumscribing a circle is a rhombus.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Marking-scheme solution
We know that lengths of tangents drawn from an external point to a circle are equal
\[\begin{aligned}
& \therefore \mathrm{AP}=\mathrm{AS} \quad ---(1) \\
& \mathrm{BP}=\mathrm{BQ} \quad ---(2) \\
& \mathrm{CR}=\mathrm{CQ} \quad ---(3) \\
& \mathrm{DR}=\mathrm{DS} \quad ---(4)
\end{aligned}
\]
Adding ($\displaystyle 1$), ($\displaystyle 2$), ($\displaystyle 3$) and ($\displaystyle 4$), we have
\[\begin{aligned}
& (\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{DR})=\mathrm{AS}+(\mathrm{BQ}+\mathrm{CQ})+\mathrm{DS} \\
& \Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{BC}+\mathrm{AD} \\
& \because \mathrm{AB}=\mathrm{CD} \text { and } \mathrm{BC}=\mathrm{AD} \\
& \therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}
\end{aligned}
\]
Therefore, ABCD is a rhombus.
PA and PB are tangents from the external point P to the circle with centre O.
\(\displaystyle \angle \mathrm{OAP}=\angle \mathrm{OBP}=90^{\circ}\)
In quadrilateral OAPB,
\[\begin{aligned}
& \angle \mathrm{APB}+\angle \mathrm{OAP}+\angle \mathrm{OBP}+\angle \mathrm{AOB}=360^{\circ} \\
& \Rightarrow \angle \mathrm{APB}+90^{\circ}+90^{\circ}+\angle \mathrm{AOB}=360^{\circ} \\
& \Rightarrow \angle \mathrm{APB}+\angle \mathrm{AOB}=180^{\circ} \\
& \therefore \angle \mathrm{APB} \text { and } \angle \mathrm{AOB} \text { are supplementary. }
\end{aligned}
\]
CirclesTangent Properties and ProofsApplyshort_answerhard
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