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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.
Solution :
Given:
Two concentric circles with a common center $O$.
Radius of the larger circle, $R = 5$ cm.
Radius of the smaller circle, $r = 3$ cm.
A chord $AB$ of the larger circle touches the smaller circle at point $P$.
To Find:
The length of the chord $AB$.
Step 1: Establishing Geometric Relationships
Let $O$ be the center of the concentric circles. Let $AB$ be the chord of the larger circle that is tangent to the smaller circle at point $P$.
By the property of tangents: A tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $OP \perp AB$. [Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.]
Step 2: Applying the Perpendicular Bisector Theorem
In the larger circle, $OP$ is a line segment from the center $O$ perpendicular to the chord $AB$.
According to the theorem: A perpendicular drawn from the center of a circle to a chord bisects the chord. [Theorem: The perpendicular from the center of a circle to a chord bisects the chord.]
Therefore, $AP = PB$.
Step 3: Calculating the Length of $AP$ using the Pythagorean Theorem
Consider the right-angled triangle $\triangle OPA$, where $\angle OPA = 90^\circ$.
Using the Pythagorean Theorem: $OA^2 = OP^2 + AP^2$.
Given $OA = R = 5$ cm and $OP = r = 3$ cm.
$5^2 = 3^2 + AP^2$
$25 = 9 + AP^2$
$AP^2 = 25 - 9$
$AP^2 = 16$
$AP = \sqrt{16} = 4$ cm.
Step 4: Determining the Total Length of the Chord $AB$
Since $AP = PB$ and $AP = 4$ cm, then $PB = 4$ cm.
The total length of the chord $AB = AP + PB$.
$AB = 4 \text{ cm} + 4 \text{ cm} = 8 \text{ cm}$.
Final Answer: The length of the chord of the larger circle is 8 cm.
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.
- 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.
- 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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