Circles — NCERT Solutions, AI Tutor & Practice

Let the two circles have centres $O$ and $O'$ with equal radii $r$. Let $AB$ be a chord in the first circle and $CD$ an equal chord in the second, so $AB = CD$.

Construction: Join $OA, OB, O'C, O'D$.

In $\triangle OAB$ and $\triangle O'CD$:

• $OA = O'C = r$ (equal radii)
• $OB = O'D = r$ (equal radii)
• $AB = CD$ (given equal chords)

By the SSS congruence rule, $\triangle OAB \cong \triangle O'CD$.

Hence by CPCT, $\angle AOB = \angle CO'D$.

Conclusion: Equal chords of congruent circles subtend equal angles at their centres.

Let $O$ be the centre with equal chords $AB = CD$. Draw $OM \perp AB$ and $ON \perp CD$.

Key fact: The perpendicular from the centre bisects the chord. So $AM = \tfrac{1}{2}AB, \quad CN = \tfrac{1}{2}CD.$

Since $AB = CD$, we get $AM = CN$.

In right triangles $\triangle OMA$ and $\triangle ONC$:

• $OA = OC = r$ (radii)
• $AM = CN$ (shown above)
• $\angle OMA = \angle ONC = 90^\circ$

By the RHS congruence rule, $\triangle OMA \cong \triangle ONC$.

By CPCT, $OM = ON$.

Conclusion: Equal chords are equidistant from the centre.

Let $O$ be the centre, $AB$ the chord of length $8$ cm, and $OM \perp AB$.

The perpendicular from the centre bisects the chord, so $AM = \tfrac{1}{2}(8) = 4 \text{ cm}.$

In right triangle $\triangle OMA$, $OA = 5$ cm (radius). By Pythagoras: $OM^2 = OA^2 - AM^2 = 5^2 - 4^2 = 25 - 16 = 9.$ $OM = 3 \text{ cm}.$

Conclusion: The chord is $3$ cm from the centre.

Radius $r = 13$ cm.

Chord of length 24 cm: half-length $= 12$ cm. $d_1 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm}.$

Chord of length 10 cm: half-length $= 5$ cm. $d_2 = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}.$

Both chords are on the same side of the centre, so the distance between them is $d_2 - d_1 = 12 - 5 = 7 \text{ cm}.$

Conclusion: The chords are $7$ cm apart.

Let $O$ be the centre and $AB$ a chord (not a diameter). Let $M$ be the midpoint of $AB$, so $AM = MB$, and $OM$ passes through the centre.

In $\triangle OMA$ and $\triangle OMB$:

• $OA = OB = r$ (radii)
• $AM = MB$ (given, $M$ is midpoint)
• $OM = OM$ (common)

By SSS congruence, $\triangle OMA \cong \triangle OMB$.

By CPCT, $\angle OMA = \angle OMB$.

But $\angle OMA + \angle OMB = 180^\circ$ (linear pair on line $AB$). So $2\angle OMA = 180^\circ \Rightarrow \angle OMA = 90^\circ$.

Conclusion: $OM \perp AB$.

Theorem: The angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining (major) arc.

Let the angle at the centre be $\angle AOB = 100^\circ$, and $\angle APB$ the angle at a point $P$ on the major arc.

$\angle APB = \tfrac{1}{2}\angle AOB = \tfrac{1}{2}(100^\circ) = 50^\circ.$

Conclusion: The angle at the major arc is $50^\circ$.

Let $AB$ be a chord and let $P, Q$ be two points on the same segment (same side of $AB$). Let $O$ be the centre.

The arc $AB$ (not containing $P, Q$) subtends $\angle AOB$ at the centre.

By the central angle theorem: $\angle APB = \tfrac{1}{2}\angle AOB \quad \text{and} \quad \angle AQB = \tfrac{1}{2}\angle AOB.$

Since both equal $\tfrac{1}{2}\angle AOB$, $\angle APB = \angle AQB.$

Conclusion: Angles in the same segment are equal.

Property: In a cyclic quadrilateral, opposite angles are supplementary (sum to $180^\circ$).

For $\angle A$ and $\angle C$: $\angle A + \angle C = 180^\circ \Rightarrow \angle C = 180^\circ - 70^\circ = 110^\circ.$

For $\angle B$ and $\angle D$: $\angle B + \angle D = 180^\circ \Rightarrow \angle D = 180^\circ - 95^\circ = 85^\circ.$

Conclusion: $\angle C = 110^\circ$ and $\angle D = 85^\circ$.

Let $AB$ be a diameter of a circle with centre $O$, and let $P$ be any point on the circle. We prove $\angle APB = 90^\circ$.

The arc $AB$ subtends $\angle AOB$ at the centre. Since $AB$ is a straight line (diameter), $\angle AOB = 180^\circ.$

By the central angle theorem, the angle at $P$ on the circle is half the central angle: $\angle APB = \tfrac{1}{2}\angle AOB = \tfrac{1}{2}(180^\circ) = 90^\circ.$

Conclusion: The angle in a semicircle is a right angle.

Let $O$ be the centre with equal chords $AB = CD$. Join $OA, OB, OC, OD$.

In $\triangle OAB$ and $\triangle OCD$:

• $OA = OC = r$ (radii)
• $OB = OD = r$ (radii)
• $AB = CD$ (given)

By SSS, $\triangle OAB \cong \triangle OCD$, so by CPCT $\angle AOB = \angle COD$.

Key fact: Arcs that subtend equal angles at the centre are equal. Since $\angle AOB = \angle COD$, $\text{arc } AB = \text{arc } CD.$

Conclusion: Equal chords cut off equal (minor) arcs.

In cyclic quadrilateral $ABCD$, $AB \parallel CD$.

Step 1 — Alternate angles: Since $AB \parallel CD$ with transversal $AC$, $\angle BAC = \angle ACD \quad (\text{alternate angles}).$

Step 2 — Equal inscribed angles ⇒ equal arcs: $\angle BAC$ is the inscribed angle standing on arc $BC$, and $\angle ACD$ stands on arc $AD$. Equal inscribed angles subtend equal arcs, so $\text{arc } BC = \text{arc } AD.$

Step 3 — Equal arcs ⇒ equal chords: $BC = AD.$

Conclusion: $AD = BC$, so $ABCD$ is an isosceles trapezium.

Let chords $AB$ and $CD$ meet at $P$ inside the circle.

Angles in the same segment: $\angle ABD$ and $\angle ACD$ both stand on arc $AD$ (same segment).

Therefore, by the 'angles in the same segment are equal' theorem: $\angle ABD = \angle ACD = 30^\circ.$

Conclusion: $\angle ABD = 30^\circ.$

(The datum $\angle APC = 40^\circ$ is extra information — the same-segment relationship alone determines $\angle ABD$.)