Exercise 10.1: Circles — NCERT Solutions

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: $OM \perp AB$.