EXAMPLES
Circles • 3 Questions
Question 1
Hint available
Prove that in two concentric , the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Key Idea
The radius drawn to the point of contact of a tangent is perpendicular to the tangent. In concentric circles the common centre lies on the perpendicular bisector of any chord of the larger circle. Hence the radius to the point of contact is the perpendicular bisector of the chord, proving the chord is bisected at that point.
Step-by-Step Solution
1. Draw two concentric circles with common centre $O$. Let the larger circle have radius $R$ and the smaller circle have radius $r$ ($r
2. Let $AB$ be a chord of the larger circle which touches the smaller circle at $T$.
3. Since $T$ is the point of contact, $OT$ is a radius of the smaller circle. By the tangent‑radius theorem, $OT\perp AB$.
4. In triangle $\triangle OAB$, we have $OA=OB=R$ (radii of the larger circle). Hence $O$ is equidistant from $A$ and $B$, so the line through $O$ perpendicular to $AB$ must be the perpendicular bisector of $AB$.
5. The line $OT$ is perpendicular to $AB$ and passes through $O$; therefore $OT$ is the perpendicular bisector of $AB$. Consequently, the foot of the perpendicular, which is the point $T$, divides $AB$ into two equal parts: $AT = TB$.
6. Hence the chord $AB$ of the larger circle is bisected at the point of contact $T$ with the smaller circle.
2. Let $AB$ be a chord of the larger circle which touches the smaller circle at $T$.
3. Since $T$ is the point of contact, $OT$ is a radius of the smaller circle. By the tangent‑radius theorem, $OT\perp AB$.
4. In triangle $\triangle OAB$, we have $OA=OB=R$ (radii of the larger circle). Hence $O$ is equidistant from $A$ and $B$, so the line through $O$ perpendicular to $AB$ must be the perpendicular bisector of $AB$.
5. The line $OT$ is perpendicular to $AB$ and passes through $O$; therefore $OT$ is the perpendicular bisector of $AB$. Consequently, the foot of the perpendicular, which is the point $T$, divides $AB$ into two equal parts: $AT = TB$.
6. Hence the chord $AB$ of the larger circle is bisected at the point of contact $T$ with the smaller circle.
Question 2
Hint available
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that PTQ = 2 OPQ.
Key Idea
Use the fact that a radius drawn to a point of tangency is perpendicular to the tangent, and that the two tangents from an external point are equal. Hence triangles \(OTP\) and \(OTQ\) are congruent, giving equal base angles. Apply the exterior angle theorem in triangle \(OPQ\).
Step-by-Step Solution
1. Draw the radii \(OP\) and \(OQ\) to the points of tangency \(P\) and \(Q\).
2. Since a radius is perpendicular to a tangent at the point of contact, we have
\[ OP \perp TP \quad \text{and} \quad OQ \perp TQ. \]
3. Hence \(\angle OPT = 90^{\circ}\) and \(\angle OQT = 90^{\circ}\).
4. From a point outside a circle, the two tangents drawn are equal; therefore
\[ TP = TQ. \]
5. Also, the radii are equal: \(OP = OQ\).
6. In triangles \(\triangle OTP\) and \(\triangle OTQ\) we have
\[ OP = OQ, \quad TP = TQ, \quad OT = OT. \]
Thus, by SSS (Side‑Side‑Side) congruence, \(\triangle OTP \cong \triangle OTQ\).
7. Consequently, the corresponding angles are equal: \(\angle POT = \angle QOT\).
8. Consider triangle \(OPQ\). The exterior angle at \(T\) for this triangle is \(\angle PTQ\). By the Exterior Angle Theorem,
\[ \angle PTQ = \angle POT + \angle QOT. \]
9. Since \(\angle POT = \angle QOT\) (from step 7), let each be \(x\). Then
\[ \angle PTQ = x + x = 2x. \]
10. But \(x = \angle OPQ\) (the angle subtended by the chord \(PQ\) at the centre). Hence
\[ \boxed{\angle PTQ = 2 \angle OPQ}. \]
2. Since a radius is perpendicular to a tangent at the point of contact, we have
\[ OP \perp TP \quad \text{and} \quad OQ \perp TQ. \]
3. Hence \(\angle OPT = 90^{\circ}\) and \(\angle OQT = 90^{\circ}\).
4. From a point outside a circle, the two tangents drawn are equal; therefore
\[ TP = TQ. \]
5. Also, the radii are equal: \(OP = OQ\).
6. In triangles \(\triangle OTP\) and \(\triangle OTQ\) we have
\[ OP = OQ, \quad TP = TQ, \quad OT = OT. \]
Thus, by SSS (Side‑Side‑Side) congruence, \(\triangle OTP \cong \triangle OTQ\).
7. Consequently, the corresponding angles are equal: \(\angle POT = \angle QOT\).
8. Consider triangle \(OPQ\). The exterior angle at \(T\) for this triangle is \(\angle PTQ\). By the Exterior Angle Theorem,
\[ \angle PTQ = \angle POT + \angle QOT. \]
9. Since \(\angle POT = \angle QOT\) (from step 7), let each be \(x\). Then
\[ \angle PTQ = x + x = 2x. \]
10. But \(x = \angle OPQ\) (the angle subtended by the chord \(PQ\) at the centre). Hence
\[ \boxed{\angle PTQ = 2 \angle OPQ}. \]
Question 3
Hint available
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig. 10.10). Find the length TP.
Key Idea
Use the relationship between the chord length and the central angle ( $PQ = 2R\sin\frac{\theta}{2}$ ), then apply the law of cosines in the isosceles triangle $\triangle PTQ$ formed by the two tangents. The tangents are equal, so $TP = TQ$, and the angle $\angle PTQ = 180^{\circ}-\theta$. This gives a solvable equation for $TP$.
Step-by-Step Solution
1. Let $O$ be the centre of the circle and $\theta = \angle POQ$ be the central angle subtended by chord $PQ$.
Using the chord‑radius formula:
$$PQ = 2R\sin\frac{\theta}{2}$$
Substituting $PQ = 8\,\text{cm}$ and $R = 5\,\text{cm}$,
$$8 = 2\times5\sin\frac{\theta}{2}\;\Rightarrow\;\sin\frac{\theta}{2}=\frac{8}{10}=0.8.$$
Hence
$$\cos\frac{\theta}{2}=\sqrt{1-\sin^{2}\frac{\theta}{2}}=\sqrt{1-0.64}=0.6.$$
Using $\cos\theta = 2\cos^{2}\frac{\theta}{2}-1$,
$$\cos\theta = 2(0.6)^{2}-1 = 0.72-1 = -0.28.$$
2. The tangents at $P$ and $Q$ meet at $T$. Since tangents from an external point are equal, let
$$TP = TQ = x\;\text{cm}.$$
The angle between the two tangents is
$$\angle PTQ = 180^{\circ}-\theta.$$
In $\triangle PTQ$ (isosceles with sides $x, x, 8$) apply the law of cosines:
$$\begin{aligned}
PQ^{2} &= x^{2}+x^{2}-2x^{2}\cos(\angle PTQ)\\
8^{2} &= 2x^{2}\bigl[1-\cos(180^{\circ}-\theta)\bigr]\\
64 &= 2x^{2}\bigl[1+\cos\theta\bigr] \quad\text{(since }\cos(180^{\circ}-\theta)=-\cos\theta\text{)}.
\end{aligned}$$
Substituting $\cos\theta = -0.28$ gives
$$64 = 2x^{2}(1-0.28)=2x^{2}(0.72)\;\Rightarrow\;x^{2}=\frac{64}{1.44}=\frac{400}{9}.$$
Hence
$$x = \sqrt{\frac{400}{9}} = \frac{20}{3}\;\text{cm}.$$
3. Therefore
$$TP = \frac{20}{3}\,\text{cm} \approx 6.67\,\text{cm}.$$
(An alternative check: using the power of a point, $TP^{2}=OT^{2}-R^{2}$, and $OT = OM+MT = 3+\frac{16}{3}=\frac{25}{3}$, which also yields $TP = \frac{20}{3}$ cm.)
Using the chord‑radius formula:
$$PQ = 2R\sin\frac{\theta}{2}$$
Substituting $PQ = 8\,\text{cm}$ and $R = 5\,\text{cm}$,
$$8 = 2\times5\sin\frac{\theta}{2}\;\Rightarrow\;\sin\frac{\theta}{2}=\frac{8}{10}=0.8.$$
Hence
$$\cos\frac{\theta}{2}=\sqrt{1-\sin^{2}\frac{\theta}{2}}=\sqrt{1-0.64}=0.6.$$
Using $\cos\theta = 2\cos^{2}\frac{\theta}{2}-1$,
$$\cos\theta = 2(0.6)^{2}-1 = 0.72-1 = -0.28.$$
2. The tangents at $P$ and $Q$ meet at $T$. Since tangents from an external point are equal, let
$$TP = TQ = x\;\text{cm}.$$
The angle between the two tangents is
$$\angle PTQ = 180^{\circ}-\theta.$$
In $\triangle PTQ$ (isosceles with sides $x, x, 8$) apply the law of cosines:
$$\begin{aligned}
PQ^{2} &= x^{2}+x^{2}-2x^{2}\cos(\angle PTQ)\\
8^{2} &= 2x^{2}\bigl[1-\cos(180^{\circ}-\theta)\bigr]\\
64 &= 2x^{2}\bigl[1+\cos\theta\bigr] \quad\text{(since }\cos(180^{\circ}-\theta)=-\cos\theta\text{)}.
\end{aligned}$$
Substituting $\cos\theta = -0.28$ gives
$$64 = 2x^{2}(1-0.28)=2x^{2}(0.72)\;\Rightarrow\;x^{2}=\frac{64}{1.44}=\frac{400}{9}.$$
Hence
$$x = \sqrt{\frac{400}{9}} = \frac{20}{3}\;\text{cm}.$$
3. Therefore
$$TP = \frac{20}{3}\,\text{cm} \approx 6.67\,\text{cm}.$$
(An alternative check: using the power of a point, $TP^{2}=OT^{2}-R^{2}$, and $OT = OM+MT = 3+\frac{16}{3}=\frac{25}{3}$, which also yields $TP = \frac{20}{3}$ cm.)