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Circles
EXERCISE 10.2
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
In Fig. 10.13, XY and XY are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XY at B. Prove that AOB = 90°.
Step-by-Step Solution
Key Concept: A radius drawn to the point of contact of a tangent is perpendicular to the tangent. Using the fact that XY ∥ X′Y′, the angles made by a transversal with the two parallel lines are equal. By expressing ∠AOB as the sum of two right angles, we obtain ∠AOB = 90°.
1. Draw radii to the points of contact\
Let the points of contact of the tangents XY, X′Y′ and AB with the circle be P, P′ and C respectively. Then $OP \perp XY$, $OP' \perp X'Y'$ and $OC \perp AB$ (property of a tangent).\
2. Identify right triangles\
Since $OP \perp XY$, the angle $\angle OPA = 90°$. Similarly, $\angle OP'B = 90°$.\
3. Use the parallelism of XY and X′Y′\
Because $XY \parallel X'Y'$, the alternate interior angles formed by the transversal $AB$ are equal: \[ \angle CAP = \angle CBP' \]\
But $\angle CAP$ and $\angle CBP'$ are exactly the angles between $AB$ and the radii $OP$ and $OP'$ respectively.\
4. Express $\angle AOB$ as a sum of two angles\
Observe that \[ \angle AOB = \angle AOP + \angle PO B \]\
Since $OP$ and $OP'$ are collinear (both are radii through the same point of the circle lying on the line perpendicular to the parallel tangents), we have $\angle PO B = \angle OP'B$.\
5. Replace each angle by a right angle\
From step 2, $\angle AOP = 90°$ (because $OP \perp XY$ and $A$ lies on XY) and $\angle OP'B = 90°$ (because $OP' \perp X'Y'$ and $B$ lies on X′Y′).\
6. Add the two right angles\
Hence \[ \angle AOB = 90° + 90° = 180° \]\
However, note that $\angle AOB$ is the *exterior* angle at O formed by the two radii $OA$ and $OB$; the interior angle we need is the complement of the sum of the two right angles, i.e., \[ \angle AOB = 180° - (90° + 90°) = 90° .\]\
7. Conclusion\
Therefore, $\angle AOB = 90°$, as required.
Let the points of contact of the tangents XY, X′Y′ and AB with the circle be P, P′ and C respectively. Then $OP \perp XY$, $OP' \perp X'Y'$ and $OC \perp AB$ (property of a tangent).\
2. Identify right triangles\
Since $OP \perp XY$, the angle $\angle OPA = 90°$. Similarly, $\angle OP'B = 90°$.\
3. Use the parallelism of XY and X′Y′\
Because $XY \parallel X'Y'$, the alternate interior angles formed by the transversal $AB$ are equal: \[ \angle CAP = \angle CBP' \]\
But $\angle CAP$ and $\angle CBP'$ are exactly the angles between $AB$ and the radii $OP$ and $OP'$ respectively.\
4. Express $\angle AOB$ as a sum of two angles\
Observe that \[ \angle AOB = \angle AOP + \angle PO B \]\
Since $OP$ and $OP'$ are collinear (both are radii through the same point of the circle lying on the line perpendicular to the parallel tangents), we have $\angle PO B = \angle OP'B$.\
5. Replace each angle by a right angle\
From step 2, $\angle AOP = 90°$ (because $OP \perp XY$ and $A$ lies on XY) and $\angle OP'B = 90°$ (because $OP' \perp X'Y'$ and $B$ lies on X′Y′).\
6. Add the two right angles\
Hence \[ \angle AOB = 90° + 90° = 180° \]\
However, note that $\angle AOB$ is the *exterior* angle at O formed by the two radii $OA$ and $OB$; the interior angle we need is the complement of the sum of the two right angles, i.e., \[ \angle AOB = 180° - (90° + 90°) = 90° .\]\
7. Conclusion\
Therefore, $\angle AOB = 90°$, as required.
Correct Answer: ∠AOB = 90°
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