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Circles
NCERT Exemplar
CBSE
Grade 10
Question:
In figure, $XY$ and $X'Y'$ are two parallel tangents to a circle with center $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^\circ$.
Step-by-Step Solution
Key Concept: Join $OC$. $\Delta OPA \cong \Delta OCA \Rightarrow \angle POA = \angle COA$. $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB$. $POQ$ is a straight line ($180^\circ$).
Join $OC$. In $\Delta OPA$ and $\Delta OCA$: $OP = OC$ (radii), $PA = CA$ (tangents), $OA = OA$ (common) $\Rightarrow \Delta OPA \cong \Delta OCA$. Thus $\angle POA = \angle COA = x$. [1.0 Mark]
Similarly, $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB = y$. [1.0 Mark]
Since $P-O-Q$ is a diameter (straight line):
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ \Rightarrow 2x + 2y = 180^\circ \Rightarrow x + y = 90^\circ$.
$\angle AOB = \angle COA + \angle COB = x + y = 90^\circ$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Proving $\Delta OPA \cong \Delta OCA \Rightarrow \angle POA = \angle COA = x$: 1.0 Mark
Proving $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB = y$: 1.0 Mark
Using straight line $2x + 2y = 180^\circ \Rightarrow \angle AOB = 90^\circ$: 1.0 Mark
Similarly, $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB = y$. [1.0 Mark]
Since $P-O-Q$ is a diameter (straight line):
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ \Rightarrow 2x + 2y = 180^\circ \Rightarrow x + y = 90^\circ$.
$\angle AOB = \angle COA + \angle COB = x + y = 90^\circ$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Proving $\Delta OPA \cong \Delta OCA \Rightarrow \angle POA = \angle COA = x$: 1.0 Mark
Proving $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB = y$: 1.0 Mark
Using straight line $2x + 2y = 180^\circ \Rightarrow \angle AOB = 90^\circ$: 1.0 Mark
Correct Answer:
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