Exercise 10.2: Circles — NCERT Solutions

Let $O$ be the centre of the circle and $P$ an external point. Let $PA$ and $PB$ be two tangents touching the circle at $A$ and $B$.

To prove: $PA = PB$.

Join $OA$, $OB$ and $OP$.

Since tangents are perpendicular to the radii at the points of contact: $\angle OAP = \angle OBP = 90^\circ.$

In triangles $OAP$ and $OBP$:

• $OA = OB$ (radii of the same circle)
• $OP = OP$ (common)
• $\angle OAP = \angle OBP = 90^\circ$

By the RHS congruence rule, $\triangle OAP \cong \triangle OBP.$

Hence by CPCT, $PA = PB.$

Thus tangents from an external point are equal in length.

Let $O$ be the common centre. Let the chord $AB$ of the larger circle (radius 5 cm) touch the smaller circle (radius 3 cm) at point $P$.

Since $AB$ is tangent to the smaller circle at $P$, $OP \perp AB, \quad OP = 3 \text{ cm}.$

The perpendicular from the centre to a chord bisects it, so $P$ is the midpoint of $AB$.

In right triangle $OPA$: $OA^2 = OP^2 + PA^2$ $5^2 = 3^2 + PA^2$ $PA^2 = 25 - 9 = 16$ $PA = 4 \text{ cm}.$

Therefore $AB = 2 \times PA = 2 \times 4 = 8 \text{ cm}.$

Let the circle touch sides $AB$, $BC$, $CD$, $DA$ at points $P$, $Q$, $R$, $S$ respectively.

Using the property that tangents from an external point are equal:

• From $A$: $AP = AS$
• From $B$: $BP = BQ$
• From $C$: $CQ = CR$
• From $D$: $DR = DS$

Add all four equations: $AP + BP + CR + DR = AS + BQ + CQ + DS$

Group the terms: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$ $AB + CD = AD + BC.$

Hence proved.

In quadrilateral $OAPB$:

• $\angle OAP = 90^\circ$ (radius $\perp$ tangent)
• $\angle OBP = 90^\circ$ (radius $\perp$ tangent)
• $\angle APB = 70^\circ$ (given)

The sum of angles of a quadrilateral is $360^\circ$: $\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ$ $\angle AOB + 90^\circ + 70^\circ + 90^\circ = 360^\circ$ $\angle AOB + 250^\circ = 360^\circ$ $\angle AOB = 110^\circ.$

Let $AB$ be a diameter of a circle with centre $O$. Let $PQ$ be the tangent at $A$ and $RS$ the tangent at $B$.

Since a tangent is perpendicular to the radius (and hence the diameter) at the point of contact: $AB \perp PQ \Rightarrow \angle PAB = 90^\circ$ $AB \perp RS \Rightarrow \angle RBA = 90^\circ$

Now $\angle PAB$ and $\angle RBA$ are alternate interior angles formed by line $AB$ cutting the two tangents. Since $\angle PAB = \angle RBA = 90^\circ,$ the alternate interior angles are equal.

Therefore $PQ \parallel RS$.

Let the incircle touch $BC$ at $D$, $CA$ at $E$, and $AB$ at $F$.

Given $BD = 8$ cm, $DC = 6$ cm.

By equal tangents:

• $BF = BD = 8$ cm
• $CE = CD = 6$ cm
• Let $AF = AE = x$

Then: $AB = AF + FB = x + 8$ $AC = AE + EC = x + 6$ $BC = BD + DC = 8 + 6 = 14$

Semi-perimeter: $s = \frac{(x+8) + (x+6) + 14}{2} = \frac{2x + 28}{2} = x + 14.$

Area using inradius $r = 4$: Area $= r \cdot s$, so $84 = 4(x + 14) \Rightarrow x + 14 = 21 \Rightarrow x = 7.$

Also verify with Heron's formula: $s = 21,\ s-a = 21-14=7,\ s-b = 21-(x+6)=8,\ s-c=21-(x+8)=6$ $\text{Area} = \sqrt{21 \times 7 \times 8 \times 6} = \sqrt{7056} = 84 \checkmark$

Therefore: $AB = x + 8 = 15 \text{ cm}, \qquad AC = x + 6 = 13 \text{ cm}.$

In triangle $OAB$, $OA = OB$ (radii), so it is isosceles and the base angles are equal: $\angle OAB = \angle OBA = 30^\circ.$

Sum of angles of triangle $OAB$: $\angle AOB = 180^\circ - 30^\circ - 30^\circ = 120^\circ.$

Now, in quadrilateral $OAPB$, the angles at $A$ and $B$ are right angles (radius $\perp$ tangent): $\angle APB + \angle AOB = 180^\circ$ $\angle APB = 180^\circ - 120^\circ = 60^\circ.$

Let $ABCD$ be a parallelogram circumscribing a circle.

Since the quadrilateral circumscribes the circle, by the tangent-length property: $AB + CD = AD + BC \quad (1)$

Since $ABCD$ is a parallelogram, opposite sides are equal: $AB = CD \quad \text{and} \quad AD = BC \quad (2)$

Substitute (2) into (1): $AB + AB = AD + AD$ $2AB = 2AD$ $AB = AD.$

So $AB = BC = CD = DA$ (using the parallelogram equalities together with $AB = AD$).

All four sides are equal, and it is already a parallelogram, so $ABCD$ is a rhombus.

Let $\angle PTQ = \theta$.

Since $TP = TQ$ (equal tangents), triangle $TPQ$ is isosceles, so its base angles are equal: $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}. \quad (1)$

Since $TP$ is a tangent and $OP$ is a radius, $\angle OPT = 90^\circ.$

Now $\angle OPT = \angle OPQ + \angle TPQ$, so $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}.$

Therefore $\angle OPQ = \frac{1}{2}\angle PTQ \quad \Rightarrow \quad \angle PTQ = 2\,\angle OPQ.$

Let the circle with centre $O$ touch $AB$, $BC$, $CD$, $DA$ at $P$, $Q$, $R$, $S$ respectively.

Join $O$ to the vertices and to the points of contact.

Since tangents from a vertex are equal and $O$ lies on the bisector of the angle at each vertex, the segment from $O$ to each vertex bisects that vertex angle. Let us denote the pairs of equal angles formed at $O$ by the tangent points:

Label the eight angles around $O$ as follows (each point of contact gives a pair): $\angle AOP = \angle AOS = a$ $\angle BOP = \angle BOQ = b$ $\angle COQ = \angle COR = c$ $\angle DOR = \angle DOS = d$

The eight angles fill up the complete angle at $O$: $2a + 2b + 2c + 2d = 360^\circ \Rightarrow a + b + c + d = 180^\circ. \quad (1)$

Now: $\angle AOB = \angle AOP + \angle POB = a + b$ $\angle COD = \angle COR + \angle ROD = c + d$

Adding: $\angle AOB + \angle COD = (a+b) + (c+d) = 180^\circ \quad \text{using } (1).$

Hence the angles subtended by opposite sides at the centre are supplementary.