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Q9:
In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.
In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.
Solution :
Given:
1. A circle with center $O$.
2. Two parallel tangents $XY$ and $X'Y'$ touching the circle at points $P$ and $Q$ respectively.
3. A third tangent $AB$ touching the circle at point $C$, intersecting $XY$ at $A$ and $X'Y'$ at $B$.
To Prove:
$\angle AOB = 90^{\circ}$
Step 1: Construction and Identification of Congruent Triangles
Join $OC$. Consider $\triangle OPA$ and $\triangle OCA$.
In $\triangle OPA$ and $\triangle OCA$:
$OP = OC$ [Radii of the same circle]
$OA = OA$ [Common side]
$AP = AC$ [Tangents drawn from an external point $A$ to the circle are equal in length]
Therefore, $\triangle OPA \cong \triangle OCA$ [By SSS Congruence Criterion].
Consequently, $\angle POA = \angle COA$ (Equation 1) [By CPCT - Corresponding Parts of Congruent Triangles].
Step 2: Applying Congruence to the Second Set of Triangles
Similarly, consider $\triangle OQB$ and $\triangle OCB$.
In $\triangle OQB$ and $\triangle OCB$:
$OQ = OC$ [Radii of the same circle]
$OB = OB$ [Common side]
$BQ = BC$ [Tangents drawn from an external point $B$ to the circle are equal in length]
Therefore, $\triangle OQB \cong \triangle OCB$ [By SSS Congruence Criterion].
Consequently, $\angle QOB = \angle COB$ (Equation 2) [By CPCT].
Step 3: Summing the Angles on the Straight Line
$PQ$ is a diameter of the circle because $XY \parallel X'Y'$ and the tangents are perpendicular to the diameter at the points of contact. Thus, $POQ$ is a straight line.
The sum of all angles formed at the center $O$ on the straight line $PQ$ is $180^{\circ}$:
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$
Step 4: Substitution and Final Calculation
Using Equation 1 ($\angle POA = \angle COA$) and Equation 2 ($\angle QOB = \angle COB$):
$\angle COA + \angle COA + \angle COB + \angle COB = 180^{\circ}$
$2\angle COA + 2\angle COB = 180^{\circ}$
$2(\angle COA + \angle COB) = 180^{\circ}$
$\angle COA + \angle COB = \frac{180^{\circ}}{2}$
$\angle AOB = 90^{\circ}$ [Since $\angle COA + \angle COB = \angle AOB$]
Final Answer: $\angle AOB = 90^{\circ}$
More Questions from Class 10 Mathematics Circles EXERCISE 10.2
- Q1: Choose the correct option and give justification. From a point $Q$, the length of the tangent to a circle is $24$ cm and the distance of $Q$ from the centre is $25$ cm. The radius of the circle is
- Q2: Choose the correct option and give justification. In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^{\circ}$, then $\angle PTQ$ is equal to
- Q3: Choose the correct option and give justification. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^{\circ}$, then $\angle POA$ is equal to
- Q4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
- Q5: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
- Q6: The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.
- Q7: Two concentric circles are of radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.
- Q8: A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$.
- Q10: 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.
- Q11: Prove that the parallelogram circumscribing a circle is a rhombus.
- Q12: A triangle $ABC$ is drawn to circumscribe a circle of radius $4$ cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8$ cm and $6$ cm respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.
- Q13: Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
CBSE Solutions for Class 10 Mathematics Circles
Chapters in CBSE - Class 10 Mathematics
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