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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.
Solution :
Given: A circle with center $O$. An external point $P$ from which two tangents $PA$ and $PB$ are drawn to the circle, touching the circle at points $A$ and $B$ respectively. The line segment $AB$ joins the points of contact.
To Prove: $\angle APB + \angle AOB = 180^\circ$ (i.e., they are supplementary).
Step 1: Identify the properties of tangents to a circle.
According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."
Therefore, $OA \perp PA$ and $OB \perp PB$.
This implies that $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
Step 2: Consider the quadrilateral $OAPB$.
In the quadrilateral $OAPB$, the sum of the interior angles is given by the angle sum property of a quadrilateral.
$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^\circ$ [Sum of interior angles of a quadrilateral is $360^\circ$].
Step 3: Substitute the known values into the equation.
Substitute $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ into the equation:
$90^\circ + \angle APB + 90^\circ + \angle AOB = 360^\circ$
Step 4: Simplify the algebraic expression.
Combine the constant terms:
$180^\circ + \angle APB + \angle AOB = 360^\circ$
Subtract $180^\circ$ from both sides of the equation:
$\angle APB + \angle AOB = 360^\circ - 180^\circ$
$\angle APB + \angle AOB = 180^\circ$
Conclusion:
Since the sum of $\angle APB$ and $\angle AOB$ is $180^\circ$, the angle between the two tangents and the angle subtended by the line segment joining the points of contact at the center are supplementary.
Final Answer: Hence, it is proved that $\angle APB + \angle AOB = 180^\circ$.
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
- 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.
- 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 $AB + CD = AD + BC$.
- 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 $\angle AOB = 90^{\circ}$.
CBSE Solutions for Class 10 Mathematics Circles
Chapters in CBSE - Class 10 Mathematics
Top Tutors who teach Circles
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