Chapter 9: Circles

Let the given circle have centre $$O$$ and radius $$r$$.

A tangent to a circle is a straight line that meets the circle in exactly one point. 
Take any point $$P$$ on the circumference. 
Draw the radius $$OP$$. 
From the theorem proved in the chapter, the tangent at an end-point of a radius is perpendicular to that radius, i.e.

$$\text{(tangent at }P) \perp OP$$

Because through a given point $$P$$ there is one and only one line that is perpendicular to a given line $$OP$$, the tangent at a fixed point $$P$$ is unique.

The circumference of a circle, however, contains infinitely many points (in fact, uncountably many). 
Since:

• there is exactly one tangent corresponding to each point on the circle, and
• there are infinitely many such points,

it follows that the circle possesses infinitely many distinct tangents.

Let O be the centre of the circle and P the point of contact of the tangent PQ. Radius $$OP = 5\,\text{cm}$$.

A fundamental theorem states that the radius drawn to the point of contact is perpendicular to the tangent, so $$OP \perp PQ$$.

Thus, in right-angled triangle $$\triangle OPQ$$,

• $$OP = 5\,\text{cm}$$ (radius)
• $$OQ = 12\,\text{cm}$$ (given)

Applying the Pythagoras theorem:

$$OQ^2 = OP^2 + PQ^2$$

$$\Rightarrow PQ^2 = OQ^2 - OP^2$$

$$\Rightarrow PQ^2 = (12)^2 - (5)^2 = 144 - 25 = 119$$

$$\Rightarrow PQ = \sqrt{119}\,\text{cm}$$

Therefore, the length of PQ is $$\sqrt{119}\,\text{cm}$$, corresponding to option (D).

Given. A straight line $$l$$.

Objective. To draw a circle and two lines that are parallel to $$l$$ such that one of them is a tangent to the circle and the other is a secant.

Construction steps

1. Draw the given line and name it $$l$$.

2. Choose any point $$O$$ not lying on $$l$$.
   Through $$O$$ draw the perpendicular $$OQ$$ to $$l$$, meeting $$l$$ at $$Q$$.
   (Use a right-angle or make arcs with a compass to erect the perpendicular.)

3. With the compass fix a length $$r$$ which is greater than $$OQ$$.
   With centre $$O$$ and radius $$r$$ draw a circle; call it $$\mathcal{C}(O,r)$$.

4. Because $$OQ<r$$, the line $$l$$ meets the circle at two distinct points; mark them $$P$$ and $$P'$$.
   Thus $$l$$ is already a secant to the circle.

5. Still keeping the same radius $$r$$ on the compass, mark a point $$T$$ on the line $$OQ$$ produced to the side of $$O$$ opposite $$Q$$ so that $$OT=r$$.

6. Through $$T$$ draw the line $$t$$ parallel to $$l$$.
   (Use set-square or the property that corresponding angles are equal.)

Justification

• The distance from the centre $$O$$ to the new line $$t$$ is the perpendicular $$OT$$, and $$OT=r$$, the radius of the circle.
• A line whose perpendicular distance from the centre equals the radius touches the circle in exactly one point; therefore $$t$$ is a tangent.
• Since $$OT=r>OQ$$, the earlier line $$l$$ intersects the circle in two points, so it is a secant.
• Both $$l$$ and $$t$$ are parallel by construction, fulfilling the requirement.

Diagram to draw: show line $$l$$ horizontally; mark $$O$$ above it; draw perpendicular $$OQ$$; draw the circle about $$O$$; mark the two points $$P,P'$$ where $$l$$ cuts the circle; mark $$T$$ on the extension of $$OQ$$ above $$O$$; through $$T$$ draw $$t$$ parallel to $$l$$ touching the circle at one point.

Given Two circles with the same centre $$O$$ (concentric circles).
Let the larger circle have radius $$R$$ and the smaller circle have radius $$r$$, where $$R > r$$.
Let $$AB$$ be a chord of the larger circle which touches the smaller circle at the point $$P$$.

To prove $$AP = PB$$; that is, the chord $$AB$$ is bisected at the point of contact $$P$$.

Construction Join $$OA,\;OB,\;OP$$.

Proof

1. Because $$AB$$ touches the smaller circle at $$P$$, it is a tangent to that circle.
   By the theorem “The radius drawn to the point of contact of a tangent is perpendicular to the tangent”, we have $$OP \perp AB$$.
2. Now consider $$AB$$ as a chord of the larger circle whose centre is also $$O$$.
   A fundamental property of a circle states: “The perpendicular drawn from the centre of a circle to a chord bisects the chord.”
   Since $$OP$$ is perpendicular to the chord $$AB$$ (from Step 1) and $$O$$ is the centre of the larger circle, it follows that $$P$$ is the midpoint of $$AB$$; hence $$AP = PB$$.

Conclusion The chord $$AB$$ of the larger circle is bisected at the point $$P$$ where it touches the smaller concentric circle. ∎

Construction Join $$OP$$, $$OQ$$ and $$PQ$$.

Step 1 Right angles at the points of contact

Since a radius is perpendicular to the tangent at the point of contact,

$$\angle OPT = 90^\circ \quad\text{and}\quad \angle OQT = 90^\circ.$$

Step 2 OPQT is a cyclic quadrilateral

In quadrilateral $$OPQT$$ the sum of the above two adjacent angles is

$$\angle OPT + \angle OQT = 90^\circ + 90^\circ = 180^\circ.$$

Therefore the opposite angles of $$OPQT$$ are supplementary, so the four points $$O, P, Q, T$$ lie on the same circle. Hence $$OPQT$$ is cyclic.

Step 3 Relating the required angles

Let

$$\angle OPQ = \theta.$$

Because $$OP = OQ$$ (radii of the given circle), triangle $$OPQ$$ is isosceles, hence

$$\angle OQP = \theta.$$

Using the angle-sum property of a triangle,

$$\angle POQ = 180^\circ - 2\theta. \quad(1)$$

Step 4 Opposite angles of the cyclic quadrilateral

In the cyclic quadrilateral $$OPQT$$, opposite angles are supplementary; therefore

$$\angle POQ + \angle PTQ = 180^\circ. \quad(2)$$

Substituting $$\angle POQ$$ from (1) in (2):

$$\bigl(180^\circ - 2\theta\bigr) + \angle PTQ = 180^\circ$$

$$\Rightarrow \angle PTQ = 2\theta.$$

But $$\theta = \angle OPQ$$, hence

$$\angle PTQ = 2\angle OPQ.$$

Hence proved.

Let O be the centre of the circle and let the perpendicular from O to the chord PQ meet PQ at M.

Because $$OP = OQ = 5\,\mathrm{cm}$$ and $$PQ = 8\,\mathrm{cm}$$, triangle OPQ is isosceles with $$OP = OQ$$.

Since $$OM \perp PQ$$, $$PM = \dfrac{PQ}{2}=4\,\mathrm{cm}$$ and $$OP = 5\,\mathrm{cm}$$.

In right ΔOPM,

$$OM^{2}=OP^{2}-PM^{2}=5^{2}-4^{2}=25-16=9\;\Rightarrow\;OM = 3\,\mathrm{cm}.$$

Let $$\angle POQ = 2\alpha$$. In right ΔOPM,

$$\sin\alpha = \dfrac{PM}{OP}= \dfrac{4}{5} \;\Longrightarrow\; \alpha = \sin^{-1}\!\left(\dfrac45\right).$$

Therefore $$\angle POQ = 2\alpha$$.

The two tangents at P and Q meet at T. The angle between them is

$$\angle PTQ = 180^{\circ}-\angle POQ = 180^{\circ}-2\alpha$$.

Because TP and TQ are tangents from the same external point T,

$$TP = TQ = x\,(\text{say}).$$

Applying the cosine rule in isosceles ΔPTQ,

$$PQ^{2}=TP^{2}+TQ^{2}-2\,(TP)(TQ)\cos\angle PTQ$$
$$\Rightarrow 8^{2}=x^{2}+x^{2}-2x^{2}\cos(180^{\circ}-2\alpha)$$
$$\Rightarrow 64 = 2x^{2}\,[1-\cos(180^{\circ}-2\alpha)].$$

Since $$\cos(180^{\circ}-\theta)=-\cos\theta$$,

$$1-\cos(180^{\circ}-2\alpha)=1+\cos 2\alpha.$$

But $$\cos 2\alpha = 1-2\sin^{2}\alpha = 1-2\left(\dfrac45\right)^{2}=1-\dfrac{32}{25}=-\dfrac{7}{25}$$, so

$$1+\cos 2\alpha = 1-\dfrac{7}{25}=\dfrac{18}{25}.$$

Substituting,

$$64 = 2x^{2}\left(\dfrac{18}{25}\right) \;\Longrightarrow\; x^{2} = \dfrac{64\times25}{36}= \dfrac{400}{9}.$$

Thus $$x = \dfrac{20}{3}\,\mathrm{cm} \approx 6.67\,\mathrm{cm}$$.

Hence $$TP = \dfrac{20}{3}\,\mathrm{cm}.$$

Given
From an external point $$Q$$ a tangent $$QP$$ is drawn to a circle with centre $$O$$.
Length of tangent: $$QP = 24\,\text{cm}$$
Distance of the point from the centre: $$OQ = 25\,\text{cm}$$
Required: radius $$OP = r$$ of the circle.

Concept used
The tangent drawn from an external point to a circle is perpendicular to the radius through the point of contact. Therefore, $$OP \perp PQ$$.

Construction
Join the centre $$O$$ to the point of contact $$P$$ and to the external point $$Q$$. This gives right △$$OQP$$ with right angle at $$P$$.

Calculation

• In right △$$OQP$$, by the Pythagoras theorem:
  $$OQ^{2} = OP^{2} + PQ^{2}$$
• Substitute the known lengths:
  $$25^{2} = r^{2} + 24^{2}$$
• Simplify:
  $$625 = r^{2} + 576$$
• Isolate $$r^{2}$$:
  $$r^{2} = 625 - 576 = 49$$
• Take the positive square root (radius is positive):
  $$r = 7\,\text{cm}$$

Conclusion
The radius of the circle is $$7\,\text{cm}$$.

Hence the correct option is (A).

Given data
From the external point T two tangents TP and TQ are drawn to a circle with centre O. The angle between the radii to the points of contact is $$\angle POQ = 110^\circ$$.

To find
The angle between the two tangents, $$\angle PTQ$$.

Step 1 – Join the centre to the external point
Join O to T. Now OP, OQ and OT are three sides of quadrilateral OPQT.

Step 2 – Use the right-angle property of a tangent
A radius is perpendicular to the tangent at the point of contact, hence
$$\angle OPT = 90^\circ$$ and $$\angle OQT = 90^\circ$$.

Step 3 – Apply angle-sum of a quadrilateral
For quadrilateral OPQT:
$$\angle OPT + \angle OQT + \angle POQ + \angle PTQ = 360^\circ$$
Substituting the known angles:
$$90^\circ + 90^\circ + 110^\circ + \angle PTQ = 360^\circ$$

Step 4 – Solve for $$\angle PTQ$$
$$\angle PTQ = 360^\circ - (90^\circ + 90^\circ + 110^\circ) = 360^\circ - 290^\circ = 70^\circ$$.

Conclusion
$$\angle PTQ = 70^\circ$$, which corresponds to option (B).

Given: From an external point $$P$$, the tangents $$PA$$ and $$PB$$ are drawn to a circle with centre $$O$$ and $$\angle APB = 80^{\circ}$$. We must find $$\angle POA$$.

1. Right angles at the points of contact
Because a radius is perpendicular to the tangent at the point of contact,
$$OA \perp PA\;\Rightarrow\;\angle OAP = 90^{\circ},$$
$$OB \perp PB\;\Rightarrow\;\angle OBP = 90^{\circ}.$$

2. Congruence of $$\triangle OPA$$ and $$\triangle OPB$$

• $$OA = OB$$ (radii)
• $$PA = PB$$ (tangents from the same external point)
• $$OP$$ is common

3. Angle at $$P$$
The ray $$OP$$ lies inside $$\angle APB$$, hence
$$\angle APB = \angle OPA + \angle OPB = \theta + \theta = 2\theta.$$
Given $$\angle APB = 80^{\circ}$$, therefore $$2\theta = 80^{\circ}\;\Rightarrow\;\theta = 40^{\circ}.$$

4. Required angle
In right triangle $$\triangle OPA$$:
$$\angle OAP = 90^{\circ},\; \angle OPA = 40^{\circ}.$$
Hence
$$\angle POA = 180^{\circ} - 90^{\circ} - 40^{\circ} = 50^{\circ}.$$

Therefore, $$\angle POA = 50^{\circ}$$, i.e. option (A).

Given: A circle with centre $$O$$ and diameter $$AB$$. Let the tangent at $$A$$ be line $$l_1$$ and the tangent at $$B$$ be line $$l_2$$.

To prove: $$l_1 \parallel l_2$$.

Construction: Join $$O$$ to $$A$$ and $$O$$ to $$B$$. Thus $$OA$$ and $$OB$$ are radii of the circle and $$AB$$ is their common straight line (the diameter).

Proof:

1. By the definition of a tangent, the tangent at any point on a circle is perpendicular to the radius drawn to that point.
   $$\Rightarrow l_1 \perp OA \qquad \text{and} \qquad l_2 \perp OB.$$
2. Since $$AB$$ is a diameter, the radii $$OA$$ and $$OB$$ lie on the same straight line; hence $$OA$$ and $$OB$$ are collinear with $$AB$$.
   Therefore $$OA$$ and $$OB$$ represent the same line $$AB$$.
3. From Steps 1 and 2: $$l_1 \perp AB \qquad \text{and} \qquad l_2 \perp AB.$$
4. Two distinct lines that are each perpendicular to the same line are parallel to each other.
   Therefore $$l_1 \parallel l_2$$.

Hence, the tangents drawn at the ends of a diameter of a circle are parallel.

Given: A circle with centre $$O$$ and a tangent $$\ell$$ that touches the circle at the point $$P$$.

To prove: The line drawn through $$P$$ perpendicular to the tangent $$\ell$$ passes through the centre $$O$$.

Construction: From the centre $$O$$ draw the perpendicular to the tangent $$\ell$$. Let the foot of this perpendicular be $$R$$, so $$OR \perp \ell$$.

Proof:

1. The point $$R$$ lies on the tangent $$\ell$$ by construction, and $$\angle ORP = 90^{\circ}$$.
2. Since $$P$$ is the only point at which $$\ell$$ meets the circle, any other point of $$\ell$$ lies outside the circle. Therefore, if $$R \neq P$$, the point $$R$$ must lie outside the circle.
3. Consider the right triangle $$\triangle ORP$$:
   $$OP^2 = OR^2 + PR^2 \quad\Rightarrow\quad OR < OP.$$ But $$OP$$ is the radius of the circle, say $$r$$, so $$OR < r$$.
4. If $$OR < r$$, the point $$R$$ is inside the circle (its distance from the centre is less than the radius). This contradicts statement 2, which said that every point of $$\ell$$ other than $$P$$ is outside the circle.
5. The contradiction arose from assuming $$R \neq P$$. Hence, $$R = P$$. Therefore the perpendicular from $$O$$ to $$\ell$$ meets $$\ell$$ exactly at $$P$$.
6. Thus the line through $$P$$ that is perpendicular to the tangent $$\ell$$ passes through the centre $$O$$.

Hence proved.

Given data

• Point A is $$5\,\text{cm}$$ away from the centre O of the circle, so $$OA = 5\,\text{cm}$$.
• The length of the tangent drawn from A touches the circle at B and is $$AB = 4\,\text{cm}$$.
• Let the radius of the circle be $$r\,\text{cm}$$, i.e. $$OB = r$$.

Reasoning using the tangent–radius property

The radius drawn to the point of contact is perpendicular to the tangent:

$$OB \perp AB$$

Therefore, $$\triangle OBA$$ is a right–angled triangle with the right angle at B.

Applying the Pythagoras theorem to $$\triangle OBA$$

For a right–angled triangle,

$$OA^{2} = OB^{2} + AB^{2}$$

Substituting the known lengths,

$$5^{2} = r^{2} + 4^{2}$$

$$25 = r^{2} + 16$$

$$r^{2} = 25 - 16$$

$$r^{2} = 9$$

$$r = 3$$ (taking the positive value, because radius is a length)

Hence, the radius of the circle is $$3\,\text{cm}$$.

Given: Two concentric circles with common centre $$O$$.

• Radius of the larger circle  $$OA = 5\,\text{cm}$$.
• Radius of the smaller circle  $$OP = 3\,\text{cm}$$.

Let $$AB$$ be a chord of the larger circle that just touches the smaller circle at point $$P$$.

To find: Length of chord $$AB$$.

Construction & Observation

• Draw radii $$OA$$ and $$OP$$. Since the two circles are concentric, $$O$$ is the common centre.
• Because $$AB$$ touches the smaller circle at $$P$$, the radius $$OP$$ is perpendicular to the tangent-chord $$AB$$; therefore $$OP \perp AB$$.
• Perpendicular from the centre to a chord bisects the chord, so $$P$$ is the midpoint of $$AB$$.
  Let $$PA = PB$$.
• Right triangle considered: $$\triangle OPA$$, right-angled at $$P$$.

Step-by-step calculation

1. Write the Pythagoras relation in $$\triangle OPA$$:
   $$OA^{2} = OP^{2} + PA^{2}$$.
2. Substitute the known lengths:
   $$5^{2} = 3^{2} + PA^{2}$$
3. Solve for $$PA$$:
   $$25 = 9 + PA^{2} \;\;\Rightarrow\;\; PA^{2} = 25 - 9 = 16$$
   $$\Rightarrow\; PA = \sqrt{16} = 4\,\text{cm}$$.
4. The full chord $$AB$$ is twice $$PA$$ (because $$P$$ is the midpoint):
   $$AB = 2 \times PA = 2 \times 4\,\text{cm} = 8\,\text{cm}$$.

Therefore, the length of the required chord is $$8\,\text{cm}$$.

Figure to draw
Draw a circle with centre O. Draw a quadrilateral ABCD around it so that the circle touches the sides AB, BC, CD, DA at points P, Q, R, S respectively. Join OP, OQ, OR, OS to the points of contact.

Goal
Prove that $$AB + CD = AD + BC$$.

Step 1 · Tangents from an external point are equal

• From point A the two tangents are AP and AS, so $$AP = AS$$.
• From point B the two tangents are BP and BQ, so $$BP = BQ$$.
• From point C the two tangents are CQ and CR, so $$CQ = CR$$.
• From point D the two tangents are DR and DS, so $$DR = DS$$.

Step 2 · Express every side in two parts

AB is made of AP and PB, therefore $$AB = AP + PB$$.
BC is made of BQ and QC, therefore $$BC = BQ + QC$$.
CD is made of CR and RD, therefore $$CD = CR + RD$$.
DA is made of DS and SA, therefore $$DA = DS + SA$$.

Step 3 · Form the required sums

The left-hand side is $$\begin{aligned} AB + CD &= (AP + PB) + (CR + RD). \end{aligned}$$

The right-hand side is $$\begin{aligned} AD + BC &= (DS + SA) + (BQ + QC). \end{aligned}$$

Step 4 · Substitute equal tangent segments

Using the equalities from Step 1:

$$AP = AS, \; BP = BQ, \; CR = CQ, \; RD = DS.$$

Replace in the two sums:

$$\begin{aligned} AB + CD &= (\underbrace{AP}_{=\,AS} + \underbrace{PB}_{=\,BQ}) + (\underbrace{CR}_{=\,CQ} + \underbrace{RD}_{=\,DS})\\[4pt] &= (AS + BQ) + (CQ + DS). \end{aligned}$$

Re-arrange the four terms:

$$AB + CD = (DS + SA) + (BQ + QC) = AD + BC.$$

Step 5 · Conclusion
Hence, for any quadrilateral circumscribing a circle, the sum of the lengths of one pair of opposite sides equals the sum of the lengths of the other pair of opposite sides, i.e. $$AB + CD = AD + BC$$. Proved.

Given data

• A circle with centre $$O$$ and radius $$r$$.
• Two parallel tangents $$XY$$ and $$X'Y'$$ touch the circle at points $$P$$ and $$Q$$ respectively.
• A third tangent $$AB$$ touches the circle at the point $$C$$ and meets $$XY$$ at $$A$$ and $$X'Y'$$ at $$B$$.

We must prove that $$\angle AOB = 90^\circ$$.

Step 1 : An analytic set–up

Choose Cartesian axes so that $$O$$ is the origin and the two parallel tangents become the horizontal lines

$$XY : y = r\quad\text{and}\quad X'Y' : y = -r$$

(The distance of each of these lines from the origin is $$|r|$$, so they are indeed tangents.)

Step 2 : Co-ordinates of the point of contact $$C$$ of the tangent $$AB$$

Let the co-ordinates of $$C$$ be $$C(x_0, y_0)$$. Because $$C$$ lies on the circle,

$$x_0^{2}+y_0^{2}=r^{2}\;\;\;\;\; (1)$$

The equation of the tangent to the circle at $$C(x_0,y_0)$$ is (Class X, Chapter 7)

$$x_0x+y_0y=r^{2}.\;\;\;\;\; (2)$$

This is precisely the straight line $$AB$$.

Step 3 : Co-ordinates of $$A$$ and $$B$$

(i) Point A

Substitute $$y=r$$ (because $$A$$ lies on $$XY$$) in (2):

$$x_0x+y_0r=r^{2}\;\Longrightarrow\;x=\dfrac{r^{2}-y_0r}{x_0}.$$

Thus

$$A\,(\,\dfrac{r^{2}-y_0r}{x_0},\;r\,).$$

(ii) Point B

Substitute $$y=-r$$ in (2):

$$x_0x-y_0r=r^{2}\;\Longrightarrow\;x=\dfrac{r^{2}+y_0r}{x_0}.$$

Hence

$$B\,(\,\dfrac{r^{2}+y_0r}{x_0},\;-r\,).$$

Step 4 : The vectors $$\overrightarrow{OA}$$ and $$\overrightarrow{OB}$$

$$\overrightarrow{OA}=\left[\dfrac{r^{2}-y_0r}{x_0},\;r\right],\qquad \overrightarrow{OB}=\left[\dfrac{r^{2}+y_0r}{x_0},\;-r\right].$$

Step 5 : Their scalar (dot) product

$$\overrightarrow{OA}\,\cdot\,\overrightarrow{OB}=\dfrac{r^{2}-y_0r}{x_0}\,\cdot\,\dfrac{r^{2}+y_0r}{x_0}+r\,(-r)$$

$$=\dfrac{r^{4}-y_0^{2}r^{2}}{x_0^{2}}-r^{2}$$

$$=\dfrac{r^{4}-y_0^{2}r^{2}-r^{2}x_0^{2}}{x_0^{2}}$$

Using relation (1), $$x_0^{2}=r^{2}-y_0^{2}$$. Substituting,

$$r^{4}-y_0^{2}r^{2}-r^{2}x_0^{2}=r^{4}-y_0^{2}r^{2}-r^{2}(r^{2}-y_0^{2})=0.$$

Therefore $$\overrightarrow{OA}\,\cdot\,\overrightarrow{OB}=0$$.

Step 6 : Geometrical conclusion

The dot product of two vectors is zero exactly when the vectors are perpendicular. Hence

$$OA \perp OB\;\;\Longrightarrow\;\;\angle AOB = 90^{\circ}.$$

Result   The required angle is a right angle.

Thus, in the configuration of Fig. 10.13, $$\angle AOB = 90^{\circ}$$ is proved.

Given : A circle with centre $$O$$. Two tangents $$PA$$ and $$PB$$ are drawn from an external point $$P$$, touching the circle at $$A$$ and $$B$$ respectively.

To prove : The angle between the tangents, $$\angle APB$$, and the angle subtended by the chord $$AB$$ at the centre, $$\angle AOB$$, are supplementary, i.e. $$\angle APB + \angle AOB = 180^{\circ}$$.

Construction : Join $$OA$$ and $$OB$$ (radii) and also join $$OP$$.

Proof :

1. Since $$PA$$ is tangent at $$A$$ and $$OA$$ is the radius through the point of contact,
   $$PA \perp OA \;\Rightarrow\; \angle OAP = 90^{\circ}$$. (Tangent ⊥ radius)
2. Similarly, $$PB$$ is tangent at $$B$$ and $$OB$$ is the radius through $$B$$,
   $$PB \perp OB \;\Rightarrow\; \angle PBO = 90^{\circ}$$.
3. Consider quadrilateral $$OAPB$$. The sum of its interior angles is $$360^{\circ}$$:
   $$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^{\circ}$$.
4. Substitute the right angles from steps 1 and 2:
   $$90^{\circ} + \angle APB + 90^{\circ} + \angle AOB = 360^{\circ}$$.
5. Simplify:
   $$\angle APB + \angle AOB = 360^{\circ} - 180^{\circ}$$
   $$\angle APB + \angle AOB = 180^{\circ}$$.

Hence, the angle between the two tangents from an external point is supplementary to the angle subtended by the line segment joining their points of contact at the centre.

Given : A parallelogram $$ABCD$$ which circumscribes a circle (the circle touches each of its four sides).

To prove : $$ABCD$$ is a rhombus, i. e. all its four sides are equal.

Construction : Let the circle touch $$AB, BC, CD, DA$$ at $$P, Q, R, S$$ respectively.
(Draw the parallelogram, inscribe the circle and mark the points of contact.)

Proof :

1. (Equal tangent segments from an external point)
   From a point outside a circle the lengths of the two tangents drawn to the circle are equal.
   Hence, writing the equal lengths from each vertex:
   • From $$A$$: $$AP = AS$$
   • From $$B$$: $$BP = BQ$$
   • From $$C$$: $$CR = CQ$$
   • From $$D$$: $$DS = DR$$
2. (Introduce symbols)
   Let
   $$AP = AS = y, bsp; BP = BQ = x, bsp; CQ = CR = z, bsp; DR = DS = w$$
3. (Express the four sides)
   $$\ AB = AP + PB = y + x$$
   $$BC = BQ + QC = x + z$$
   $$CD = CR + RD = z + w$$
   $$DA = DS + SA = w + y$$
4. (Use properties of a parallelogram)
   In any parallelogram, opposite sides are equal, so $$AB = CD$$ and $$BC = DA$$. Substituting from step 3:
   $$y + x = z + w \quad(1)$$ $$x + z = w + y \[2]$$
5. (Solve the two equations)
   Subtract $$[2]$$ from $$[1]$$: $$ (y + x) - (x + z) = (z + w) - (w + y) $$
   $$y - z = z - y$$
   $$2y = 2z \;\Rightarrow\; y = z$$
6. Put $$y = z$$ into $$[2]$$: $$x + y = w + y \;\Rightarrow\; x = w$$
7. (All four sides are equal)
   $$\ AB = y + x$$
   $$BC = x + z = x + y = AB$$
   $$CD = z + w = y + x = AB$$
   $$DA = w + y = x + y = AB$$
   Therefore $$AB = BC = CD = DA$$.

All four sides of $$ABCD$$ are equal, while opposite sides are parallel by definition of a parallelogram; hence $$ABCD$$ is a rhombus.

Hence, a parallelogram that circumscribes a circle is always a rhombus.

The circle is the incircle of triangle ABC, touching BC at D, AB at F and AC at E.

1. Equal tangent segments
From an external point the two tangents to a circle have the same length, hence

• From B: $$BD = BF = 8\,\text{cm}$$
• From C: $$CD = CE = 6\,\text{cm}$$
• From A: $$AF = AE = x\,\text{cm} \;\; (\text{say})$$

2. Express the three sides of the triangle

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

3. Semiperimeter
$$s = \tfrac12(AB + BC + AC) = \tfrac12 \big[(x+8) + 14 + (x+6)\big] = \tfrac12(2x + 28) = x + 14$$

4. Area via inradius
The inradius is given as $$r = 4\,\text{cm}$$, so

$$\Delta = r\,s = 4(x + 14) = 4x + 56$$

5. Area via Heron’s formula

$$\Delta = \sqrt{s\,(s-AB)\,(s-BC)\,(s-AC)}$$

Compute each factor:

• $$s-AB = (x+14) - (x+8) = 6$$
• $$s-BC = (x+14) - 14 = x$$
• $$s-AC = (x+14) - (x+6) = 8$$

Hence

$$\Delta = \sqrt{(x+14)\,(x)\,(8)\,(6)} = \sqrt{48x(x+14)}$$

6. Equate the two expressions for area

$$\sqrt{48x(x+14)} = 4x + 56$$

Square both sides:

$$48x(x+14) = (4x + 56)^2$$

Notice $$4x + 56 = 4(x+14)$$, therefore

$$48x(x+14) = 16(x+14)^2$$

Divide by $$16(x+14) \,(\neq 0)$$:

$$3x = x + 14$$

$$2x = 14 \;\;\Longrightarrow\;\; x = 7$$

7. Required sides

• $$AB = x + 8 = 7 + 8 = 15\,\text{cm}$$
• $$AC = x + 6 = 7 + 6 = 13\,\text{cm}$$

The triangle sides are $$AB = 15\,\text{cm},\; AC = 13\,\text{cm},\; BC = 14\,\text{cm}$$, satisfying the triangle inequality, so the construction is possible.

To prove: If a circle with centre O is tangent to all the four sides of quadrilateral ABCD, then the angles subtended at O by each pair of opposite sides are supplementary, i.e. $$\angle AOB + \angle COD = 180^{\circ}$$ and $$\angle BOC + \angle DOA = 180^{\circ}$$.

Figure to draw: A convex quadrilateral ABCD. Draw the circle that touches AB, BC, CD, DA at points P, Q, R, S respectively; mark its centre O. Join OA, OB, OC, OD.

Reasoning steps

1. Because the circle is tangent to sides AB and AD at P and S, the distances $$d(O,AB)=d(O,AD)$$. Hence O lies on the bisector of $$\angle DAB$$. Therefore $$OA$$ is the internal angle-bisector of $$\angle DAB$$.
   Exactly the same argument shows that

   • $$OB$$ bisects $$\angle ABC$$,
   • $$OC$$ bisects $$\angle BCD$$,
   • $$OD$$ bisects $$\angle CDA$$.

2. Consider $$\triangle AOB$$.
   Because $$OA$$ bisects $$\angle DAB$$, we have $$\angle OAB = \tfrac12\angle DAB$$; because $$OB$$ bisects $$\angle ABC$$, we have $$\angle OBA = \tfrac12\angle ABC$$.

   Applying the angle-sum property of a triangle,

   $$\angle AOB + \angle OAB + \angle OBA = 180^{\circ}$$

   $$\Rightarrow \angle AOB + \tfrac12\angle DAB + \tfrac12\angle ABC = 180^{\circ}$$

   $$\Rightarrow \angle AOB = 180^{\circ} - \tfrac12(\angle DAB + \angle ABC).$$

3. In exactly the same way, for $$\triangle COD$$ we have

   $$\angle COD = 180^{\circ} - \tfrac12(\angle CDA + \angle BCD).$$

4. Add the two expressions obtained in Steps 2 and 3:

   $$\angle AOB + \angle COD = \bigl[180^{\circ} - \tfrac12(\angle DAB + \angle ABC)\bigr] + \bigl[180^{\circ} - \tfrac12(\angle CDA + \angle BCD)\bigr].$$

   Simplify:

   $$\angle AOB + \angle COD = 360^{\circ} - \tfrac12(\angle DAB + \angle ABC + \angle CDA + \angle BCD).$$

5. For any quadrilateral, the sum of its interior angles is $$360^{\circ}$$, i.e.

   $$\angle DAB + \angle ABC + \angle BCD + \angle CDA = 360^{\circ}.$$

   Substituting in the expression from Step 4:

   $$\angle AOB + \angle COD = 360^{\circ} - \tfrac12(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}.$$

6. Exactly the same chain of arguments, applied to triangles $$BOC$$ and $$DOA$$, gives

   $$\angle BOC + \angle DOA = 180^{\circ}.$$

Conclusion: Each pair of opposite sides of a quadrilateral that circumscribes a circle forms supplementary angles at the centre of that circle. Hence proved.