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Q5:
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution :
Given: A circle with center $O$. Let $XY$ be a tangent to the circle at the point of contact $P$.
To Prove: The perpendicular to the tangent $XY$ at the point of contact $P$ passes through the center $O$.
Step 1: Assumption for Proof by Contradiction
Let us assume that the perpendicular to the tangent $XY$ at point $P$ does not pass through the center $O$. Instead, let it pass through another point $O'$ such that $O'P \perp XY$.
Step 2: Utilizing the Tangent-Radius Theorem
We know from the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."
Therefore, the radius $OP$ must be perpendicular to the tangent $XY$.
Mathematically, $OP \perp XY$.
This implies that $\angle OPX = 90^\circ$. [Since $OP$ is the radius at point of contact $P$]
Step 3: Analyzing the Assumption
By our initial assumption in Step 1, we stated that $O'P \perp XY$.
This implies that $\angle O'PX = 90^\circ$.
Step 4: Logical Deduction
Comparing the results from Step 2 and Step 3:
$\angle OPX = 90^\circ$
$\angle O'PX = 90^\circ$
Therefore, $\angle OPX = \angle O'PX$.
Step 5: Conclusion
The equality $\angle OPX = \angle O'PX$ is only possible if the line $O'P$ coincides with the line $OP$.
This means that the point $O'$ must be the same as point $O$.
Thus, our assumption that the perpendicular does not pass through the center is false.
Final Answer: Hence, it is proved that the perpendicular at the point of contact to the tangent to a circle must pass through the center.
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
- 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.
- 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.
- 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
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