Examples - Chapter 10 Circles class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 10 - Circles - Ncert Solutions class 10 - Maths

Example 1 :

Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Solution:

We are given two concentric circles $C_1$ and $C_2$ with centre $O$ and a chord $A B$ of the larger circle $\mathrm{C}_1$ which touches the smaller circle $\mathrm{C}_2$ at the point $\mathrm{P}$ (see Fig. 10.8). We need to prove that $\mathrm{AP}=\mathrm{BP}$.

Let us join OP. Then, $A B$ is a tangent to $C_2$ at $P$ and $\mathrm{OP}$ is its radius. Therefore, by Theorem 10.1,
$
\mathrm{OP} \perp \mathrm{AB}
$

Now $A B$ is a chord of the circle $C_1$ and $O P \perp A B$. Therefore, $O P$ is the bisector of the chord $A B$, as the perpendicular from the centre bisects the chord,
i.e.,
$
\mathrm{AP}=\mathrm{BP}
$

Now $A B$ is a chord of the circle $C_1$ and $O P \perp A B$. Therefore, $O P$ is the bisector of the chord $A B$, as the perpendicular from the centre bisects the chord,
i.e.,
$
A P=B P
$

Example 2 :

Two tangents TP and TQ are drawn to a circle with centre $\mathrm{O}$ from an external point $\mathrm{T}$. Prove that $\angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}$.

Solution:

We are given a circle with centre $\mathrm{O}$, an external point $T$ and two tangents $T P$ and $T Q$ to the circle, where $\mathrm{P}, \mathrm{Q}$ are the points of contact (see Fig. 10.9). We need to prove that
$
\angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}
$

Let
$
\angle \mathrm{PTQ}=\theta
$

Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle.
Therefore,
$
\angle \mathrm{TPQ}=\angle \mathrm{TQP}=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{1}{2} \theta
$

Also, by Theorem 10.1,
$
\angle \mathrm{OPT}=90^{\circ}
$

So,
$
\begin{aligned}
\angle \mathrm{OPQ}=\angle \mathrm{OPT}-\angle \mathrm{TPQ} & =90^{\circ}-\left(90^{\circ}-\frac{1}{2} \theta\right) \\
& =\frac{1}{2} \theta=\frac{1}{2} \angle \mathrm{PTQ} \\
\angle \mathrm{PTQ} & =2 \angle \mathrm{OPQ}
\end{aligned}
$

This gives
$
\begin{aligned}
& =\frac{1}{2} \theta=\frac{1}{2} \angle \mathrm{PTQ} \\
\angle \mathrm{PTQ} & =2 \angle \mathrm{OPQ}
\end{aligned}
$

Example 3:

$\mathrm{PQ}$ is a chord of length $8 \mathrm{~cm}$ of a circle of radius $5 \mathrm{~cm}$. The tangents at $\mathrm{P}$ and $\mathrm{Q}$ intersect at a point $\mathrm{T}$ (see Fig. 10.10). Find the length TP.

Solution :

Join OT. Let it intersect PQ at the point R. Then $\Delta \mathrm{TPQ}$ is isosceles and TO is the angle bisector of $\angle \mathrm{PTQ}$. So, OT $\perp \mathrm{PQ}$ and therefore, OT bisects PQ which gives $\mathrm{PR}=\mathrm{RQ}=4 \mathrm{~cm}$.

Also, $\mathrm{OR}=\sqrt{\mathrm{OP}^2-\mathrm{PR}^2}=\sqrt{5^2-4^2} \mathrm{~cm}=3 \mathrm{~cm}$.

Now, $\angle \mathrm{TPR}+\angle \mathrm{RPO}=90^{\circ}=\angle \mathrm{TPR}+\angle \mathrm{PTR}$
(Why?)

So, $\angle \mathrm{RPO}=\angle \mathrm{PTR}$
Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.

This gives
$
\frac{\mathrm{TP}}{\mathrm{PO}}=\frac{\mathrm{RP}}{\mathrm{RO}} \text {, i.e., } \frac{\mathrm{TP}}{5}=\frac{4}{3} \text { or } \mathrm{TP}=\frac{20}{3} \mathrm{~cm} \text {. }
$

Note : TP can also be found by using the Pythagoras Theorem, as follows:
Let
$
\begin{array}{rlrl}
\mathrm{TP} & =x \text { and } \mathrm{TR}=y . \quad \text { Then } \\
x^2 & =y^2+16 & & \text { (Taking right } \Delta \mathrm{PRT}) \\
x^2+5^2 & =(y+3)^2 & & (\text { Taking right } \Delta \mathrm{OPT})
\end{array}
$

Subtracting (1) from (2), we get
$
25=6 y-7 \text { or } y=\frac{32}{6}=\frac{16}{3}
$

Therefore,
$
x^2=\left(\frac{16}{3}\right)^2+16=\frac{16}{9}(16+9)=\frac{16 \times 25}{9} \quad[\text { From (1)] }
$
or
$
x=\frac{20}{3}
$