End-of-Chapter Exercises

Let the chord be $AB$, and let $OM$ be the perpendicular from center $O$ to $AB$.
• Radius $r = OA = 13\text{ cm}$.
• Perpendicular distance $d = OM = 5\text{ cm}$.

In right-angled triangle $\Delta OMA$, by Baudhāyana-Pythagoras theorem:
$$OA^2 = OM^2 + AM^2 \implies 13^2 = 5^2 + AM^2$$
$$169 = 25 + AM^2 \implies AM^2 = 144 \implies AM = 12\text{ cm}$$

Since the perpendicular from the centre bisects the chord:
$$\text{Chord Length } AB = 2 \times AM = 2 \times 12 = \mathbf{24\text{ cm}}$$

By the central angle theorem: "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle."

Let $\theta$ be the angle subtended at the circumference:
$$\text{Central Angle} = 2\theta \implies 70^\circ = 2\theta \implies \theta = \frac{70^\circ}{2} = \mathbf{35^\circ}$$

• Radius $r = \text{Diameter} / 2 = 26 / 2 = 13\text{ cm}$.
• Perpendicular from the centre bisects the chord, so the half-chord length is $24 / 2 = 12\text{ cm}$.

Using the Baudhāyana-Pythagoras theorem in right triangle $\Delta OMA$:
$$r^2 = d^2 + (\text{half-chord})^2 \implies 13^2 = d^2 + 12^2$$
$$169 = d^2 + 144 \implies d^2 = 25 \implies d = \mathbf{5\text{ cm}}$$

• Radius $r = 15\text{ cm}$.
• Perpendicular distance $d = 9\text{ cm}$.

Using Baudhāyana-Pythagoras theorem in the right triangle:
$$\text{Half-chord length } x = \sqrt{r^2 - d^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm}$$

$$\text{Total Chord Length} = 2x = 2 \times 12 = \mathbf{24\text{ cm}}$$

Let $AB$ be a chord. Let line $L$ be the perpendicular bisector of $AB$, intersecting $AB$ at its midpoint $M$.
Let $O$ be the centre of the circle. Join $OA$ and $OB$.

Compare triangles $\Delta OMA$ and $\Delta OMB$:
1. $OA = OB$ (radii of circle)
2. $AM = BM$ ($M$ is midpoint of $AB$)
3. $OM = OM$ (common side)

By SSS congruence criterion, $\Delta OMA \cong \Delta OMB$.
This implies:
$$\angle OMA = \angle OMB$$

Since $A-M-B$ is a straight line, $\angle OMA$ and $\angle OMB$ form a linear pair:
$$\angle OMA + \angle OMB = 180^\circ \implies 2\angle OMA = 180^\circ \implies \angle OMA = 90^\circ$$
This proves that the line segment joining the centre $O$ to the midpoint $M$ is perpendicular to $AB$. Thus, the perpendicular bisector $L$ must pass through $O$.

Since $AB$ is the diameter, the arc $ACB$ forms a semicircle.
• The central angle subtended by the diameter $AB$ at the centre $O$ is a straight line angle, which is exactly $180^\circ$.

By the central angle theorem, the angle subtended by the arc at any point $C$ on the circumference is half of the central angle:
$$\angle ACB = \frac{1}{2} \times 180^\circ = \mathbf{90^\circ}$$
Therefore, the measure of $\angle ACB$ is $90^\circ$ (angle in a semicircle is a right angle).

By the cyclic quadrilateral theorem, opposite angles of a cyclic quadrilateral are supplementary (their sum is $180^\circ$).

$$\angle A + \angle C = 180^\circ \implies 75^\circ + \angle C = 180^\circ \implies \angle C = \mathbf{105^\circ}$$

$$\angle B + \angle D = 180^\circ \implies 110^\circ + \angle D = 180^\circ \implies \angle D = \mathbf{70^\circ}$$

Since $PQRS$ is inscribed in a circle, it is a cyclic quadrilateral. The opposite angles $\angle P$ and $\angle R$ are supplementary:
$$\angle P + \angle R = 180^\circ$$
$$(2x + 10) + (3x - 20) = 180$$

$$5x - 10 = 180 \implies 5x = 190 \implies x = \mathbf{38}$$

Now calculate the angles:
• $\angle P = 2(38) + 10 = 76 + 10 = \mathbf{86^\circ}$
• $\angle R = 3(38) - 20 = 114 - 20 = \mathbf{94^\circ}$

• Perpendicular distance $d = 6\text{ cm}$.
• Half-chord length $x = 16 / 2 = 8\text{ cm}$.

Using the Baudhāyana-Pythagoras theorem in right triangle:
$$\text{Radius } r = \sqrt{d^2 + x^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10\text{ cm}}$$

Let the sides of the cyclic quadrilateral be $a=5, b=5, c=12, d=12$.
Use **Brahmagupta's formula**:
$$\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$
where $s$ is the semi-perimeter:
$$s = \frac{a+b+c+d}{2} = \frac{5+5+12+12}{2} = \frac{34}{2} = 17$$

Calculate the area:
$$\text{Area} = \sqrt{(17-5)(17-5)(17-12)(17-12)} = \sqrt{12 \times 12 \times 5 \times 5} = 12 \times 5 = \mathbf{60\text{ sq. units}}$$

Let the cyclic quadrilateral be $ABCD$ inscribed in a circle with center $O$.
• The center $O$ lies **inside** the quadrilateral if and only if all four arcs subtended by the sides ($AB, BC, CD, DA$) are minor arcs (i.e. each central angle is $< 180^\circ$).

• **Best check using angle criterion**:
Let $P$ be the intersection of the diagonals. If the perpendicular projections of $O$ onto all sides lie within the boundaries of the sides, $O$ is inside.
Alternatively, if any side subtends an obtuse angle at the circumference that is greater than $90^\circ$ (which means its central angle $> 180^\circ$), the centre $O$ must lie **outside** the quadrilateral. Since opposite angles sum to $180^\circ$, if a diagonal divides it into two triangles and both are acute, the centre lies inside.

Let $AB$ and $CD$ be two equal chords of a circle with centre $O$, intersecting at point $P$.
• Draw perpendiculars $OM \perp AB$ and $ON \perp CD$. Since equal chords are equidistant from the centre:
$$OM = ON$$

In right-angled triangles $\Delta OMP$ and $\Delta ONP$:
1. $\angle OMP = \angle ONP = 90^\circ$ (perpendicularity)
2. $OP = OP$ (common hypotenuse)
3. $OM = ON$ (equal chord distance)
By RHS congruence: $\Delta OMP \cong \Delta ONP \implies MP = NP$.

Since the perpendicular from the centre bisects the chords:
• $AM = MB = \frac{AB}{2}$
• $CN = ND = \frac{CD}{2}$
Since $AB = CD \implies AM = CN$ and $MB = ND$.

Now compute corresponding segments:
• $AP = AM + MP = CN + NP = CP \implies \mathbf{AP = CP}$
• $BP = MB - MP = ND - NP = DP \implies \mathbf{BP = DP}$
This proves that the segments of one chord are equal to the corresponding segments of the other.

Let chord be $AB = 6\text{ cm}$ and distance $OM = 3\text{ cm}$ where $M$ is midpoint of $AB$.
• Half-chord $AM = 3\text{ cm}$.
• Perpendicular distance $OM = 3\text{ cm}$.

In right triangle $\Delta OMA$, since $AM = OM = 3\text{ cm}$, it is an isosceles right-angled triangle.
• $\angle AOM = \angle OAM = 45^\circ$.
• Radius $OA = \sqrt{OM^2 + AM^2} = \sqrt{3^2 + 3^2} = \mathbf{3\sqrt{2}\text{ cm} \approx 4.24\text{ cm}}$.

This circle is the circumcircle of an isosceles right-angled triangle $\Delta OAB$ where the central angle $\angle AOB = 90^\circ$ and radius is $3\sqrt{2}\text{ cm}$.

Let $ABCD$ be a parallelogram inscribed in a circle.
• Since it is inscribed in a circle, $ABCD$ is a cyclic quadrilateral.
• Therefore, opposite angles sum to $180^\circ$:
$$\angle A + \angle C = 180^\circ$$

• In any parallelogram, opposite angles are equal:
$$\angle A = \angle C$$

Substitute $\angle C = \angle A$ into the sum equation:
$$2\angle A = 180^\circ \implies \angle A = 90^\circ$$
Since a parallelogram with one right angle is a rectangle, all angles of $ABCD$ are $90^\circ$.
Thus, $ABCD$ is a rectangle.

Let $ABCD$ be a rectangle inscribed in a circle.
• The diagonals $AC$ and $BD$ of a rectangle are equal and bisect each other at a point $P$.

• Since $\angle ABC = 90^\circ$, $AC$ subtends a right angle on the circumference.
• By theorem, the chord subtending $90^\circ$ on the circumference must be a diameter of the circle. Thus, diagonal $AC$ is a diameter.
• Similarly, diagonal $BD$ is also a diameter.

• The diameters of a circle always intersect at the centre of the circle.
• Therefore, the diagonal intersection point $P$ must be the centre of the circle.

Let the fixed length of the chords be $l$.
• By theorem, all chords of equal length in a circle are equidistant from the centre.
• This means the perpendicular distance $d$ of the midpoint of each chord from the centre $O$ is constant:
$$d = \sqrt{r^2 - \left(\frac{l}{2}\right)^2}$$

The set of all points that are at a constant distance $d$ from the fixed centre $O$ forms a **circle**.
Therefore, the shape formed by the midpoints is a concentric circle of radius $d$.

Join the centre $O$ to vertices $A$, $B$, and $C$.
Compare triangles $\Delta OAB$ and $\Delta OAC$:
1. $AB = AC$ (given congruent chords)
2. $OA = OA$ (common side)
3. $OB = OC$ (radii of circle)

By SSS congruence criterion, $\Delta OAB \cong \Delta OAC$.
This implies:
$$\angle OAB = \angle OAC$$

Since $\angle OAB = \angle OAC$, the line segment $AO$ bisects the angle $\angle BAC$.
Therefore, the centre $O$ lies on the angle bisector of $\angle BAC$.

Let $AB = 10\text{ cm}$ and $CD = 24\text{ cm}$ be the parallel chords. Let $O$ be the centre and $r$ be the radius.
Draw perpendicular from $O$ meeting $AB$ at $M$ and $CD$ at $N$.
• $AM = 10 / 2 = 5\text{ cm}$.
• $CN = 24 / 2 = 12\text{ cm}$.
Let $ON = x$. Since the distance between chords is $7\text{ cm}$, $OM = x + 7$.

In right triangle $\Delta ONC$:
$$r^2 = ON^2 + CN^2 = x^2 + 12^2 = x^2 + 144 \quad \text{--- (Equation 1)}$$

In right triangle $\Delta OMA$:
$$r^2 = OM^2 + AM^2 = (x + 7)^2 + 5^2 = x^2 + 14x + 49 + 25 = x^2 + 14x + 74 \quad \text{--- (Equation 2)}$$

Equate Equation 1 and Equation 2:
$$x^2 + 144 = x^2 + 14x + 74 \implies 14x = 70 \implies x = 5\text{ cm}$$

Substitute $x = 5$ into Equation 1:
$$r^2 = 5^2 + 144 = 25 + 144 = 169 \implies r = \mathbf{13\text{ cm}}$$

(i) Side length:
A regular hexagon inscribed in a circle consists of 6 congruent equilateral triangles with vertices at the centre.
Therefore, the side length $s$ of the hexagon is equal to the radius:
$$s = \mathbf{r}$$

(ii) Distance from the centre:
This distance $d$ is the altitude of one of the equilateral triangles of side $r$.
$$d = r\sin(60^\circ) = \mathbf{\frac{\sqrt{3}}{2} r \approx 0.866 r}$$

Since the points $M, N, O, P$ lie on the circle, the segments $MP$ forms a chord.
• The angles $\angle MOP$ and $\angle MNP$ have their vertices $O$ and $N$ on the circumference.
• Both angles are subtended by the same chord $MP$ on the same segment of the circle.

By the theorem: "Angles subtended by an arc in the same segment of a circle are equal."
$$\mathbf{\angle MOP = \angle MNP}$$

• In cyclic quadrilateral $ABCD$, opposite angles are supplementary:
$$\angle ABC + \angle ADC = 180^\circ \quad \text{--- (Equation 1)}$$

• Since $E$ lies on the extension of $CD$, the points $C, D, E$ lie on a straight line. The angles $\angle ADC$ and $\angle CDE$ form a linear pair:
$$\angle ADC + \angle CDE = 180^\circ \quad \text{--- (Equation 2)}$$

Equate Equation 1 and Equation 2:
$$\angle ABC + \angle ADC = \angle ADC + \angle CDE$$
Subtract $\angle ADC$ from both sides:
$$\mathbf{\angle CDE = \angle ABC}$$

Let $AB$ be a chord of a circle with centre $O$ and radius $r$. Let $d$ be its perpendicular distance from the centre.
• The length of the chord is:
$$AB = 2\sqrt{r^2 - d^2}$$

• The distance $d$ must satisfy $d^2 \ge 0$.
• To maximize $AB$, we must minimize $d^2$. The smallest value of $d^2$ is $0$, which occurs when the chord passes directly through the centre ($d = 0$):
$$AB_{max} = 2\sqrt{r^2 - 0} = 2r = \text{Diameter}$$
• For any other chord, $d > 0 \implies AB < 2r$.

Let $CD$ be any chord passing through $A$, and let $OM$ be the perpendicular from centre $O$ to $CD$.
• In right-angled triangle $\Delta OMA$, the hypotenuse is $OA$. Thus, we always have:
$$OM \le OA$$

• The length of the chord is $CD = 2\sqrt{r^2 - OM^2}$.
• To minimize the chord length, we must maximize the perpendicular distance $OM$.
• Since $OM \le OA$, the maximum value of $OM$ is when $OM = OA$.

• This maximum is achieved when $M$ coincides with $A$, meaning the perpendicular from $O$ to the chord meets it exactly at $A$.
• Hence, $OA \perp CD$, and the shortest chord is perpendicular to $OA$.

Let the diameter be $BC$ with centre $O$, and $A$ be a point on the semicircle.
• Draw $OA$. The segments $OA$, $OB$, and $OC$ are radii:
$$OA = OB = OC = r$$

• In isosceles triangle $\Delta OAB$ (since $OA = OB$):
$$\angle OAB = \angle OBA = a$$
• In isosceles triangle $\Delta OAC$ (since $OA = OC$):
$$\angle OAC = \angle OCA = b$$

• The total angle at vertex $A$ is:
$$\angle BAC = a + b$$
• Using the angle sum property in $\Delta ABC$:
$$\angle B + \angle C + \angle BAC = 180^\circ$$
$$a + b + (a + b) = 180^\circ \implies 2(a + b) = 180^\circ \implies a + b = \mathbf{90^\circ}$$
Thus, $\angle BAC = 90^\circ$, proving the statement.

Let the circle be centered at origin $O$, and let diameter $AB$ lie along the horizontal x-axis.
• Since chords $CC'$ and $DD'$ are perpendicular to $AB$ (the x-axis), they are vertical lines.
• Thus, the points are symmetric across the x-axis:
$$C = (x_1, y_1), \quad C' = (x_1, -y_1)$$
$$D = (x_2, y_2), \quad D' = (x_2, -y_2)$$

Calculate midpoints $M$ (midpoint of $CD$) and $M'$ (midpoint of $C'D'$):
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
$$M' = \left(\frac{x_1 + x_2}{2}, \frac{-y_1 - y_2}{2}\right)$$

Since the x-coordinates of $M$ and $M'$ are equal ($x_M = x_{M'} = \frac{x_1+x_2}{2}$), the line joining them is vertical.
• Since $AB$ lies on the horizontal x-axis, any vertical line is perpendicular to $AB$.
• Thus, $MM' \perp AB$.

Let $ABCD$ be the cyclic quadrilateral with circumcentre $O$.
• Join $OA$, $OB$, $OC$, and $OD$. These segment are equal radii ($OA=OB=OC=OD=r$), dividing the quadrilateral into 4 isosceles triangles.

Let the base angles of these triangles be:
• In $\Delta OAB$: $\angle OAB = \angle OBA = q$
• In $\Delta OBC$: $\angle OBC = \angle OCB = u$
• In $\Delta OCD$: $\angle OCD = \angle ODC = v$
• In $\Delta ODA$: $\angle ODA = \angle OAD = p$

The interior angles of quadrilateral $ABCD$ are:
$$\angle A = p + q, \quad \angle B = q + u, \quad \angle C = u + v, \quad \angle D = v + p$$

Sum of all interior angles of a quadrilateral is $360^\circ$:
$$\angle A + \angle B + \angle C + \angle D = 360^\circ$$
$$(p + q) + (q + u) + (u + v) + (v + p) = 360^\circ$$
$$2(p + q + u + v) = 360^\circ \implies p + q + u + v = 180^\circ$$

Now compute sum of opposite angles:
$$\angle A + \angle C = (p + q) + (u + v) = p + q + u + v = \mathbf{180^\circ}$$
$$\angle B + \angle D = (q + u) + (v + p) = p + q + u + v = \mathbf{180^\circ}$$
This justifies the theorem.