Circles — NCERT Solutions

Given: Two congruent circles with centres O and O′ and equal radii r. AB and CD are equal chords of the two circles respectively, i.e., AB = CD.

To Prove: ∠AOB = ∠CO′D

Proof:

In △AOB and △CO′D:

$$OA = O'C = r \quad \text{(radii of congruent circles)}$$

$$OB = O'D = r \quad \text{(radii of congruent circles)}$$

$$AB = CD \quad \text{(given)}$$

By SSS congruence criterion:
$$\triangle AOB \cong \triangle CO'D$$

Therefore, by CPCT:
$$\angle AOB = \angle CO'D$$

Hence proved. Equal chords of congruent circles subtend equal angles at their centres.

Given: Two congruent circles with centres O and O′ and equal radii r. Chords AB and CD subtend equal angles at their respective centres, i.e., ∠AOB = ∠CO′D.

To Prove: AB = CD

Proof:

In △AOB and △CO′D:

$$OA = O'C = r \quad \text{(radii of congruent circles)}$$

$$OB = O'D = r \quad \text{(radii of congruent circles)}$$

$$\angle AOB = \angle CO'D \quad \text{(given)}$$

By SAS congruence criterion:
$$\triangle AOB \cong \triangle CO'D$$

Therefore, by CPCT:
$$AB = CD$$

Hence proved. If chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Given: Two circles with centres O and O′, radii 5 cm and 3 cm respectively. They intersect at points A and B. Distance OO′ = 4 cm.

To Find: Length of common chord AB.

Solution:

Let the common chord AB intersect OO′ at point M.

Let OM = x, so O′M = 4 − x.

In △OMA (right-angled at M, since the line joining centres is perpendicular to the common chord):
$$OA^2 = OM^2 + AM^2$$
$$25 = x^2 + AM^2 \quad \cdots (1)$$

In △O′MA (right-angled at M):
$$O'A^2 = O'M^2 + AM^2$$
$$9 = (4-x)^2 + AM^2 \quad \cdots (2)$$

Subtracting (2) from (1):
$$25 - 9 = x^2 - (4-x)^2$$
$$16 = x^2 - 16 + 8x - x^2$$
$$16 = 8x - 16$$
$$8x = 32$$
$$x = 4 \text{ cm}$$

Substituting in (1):
$$AM^2 = 25 - 16 = 9$$
$$AM = 3 \text{ cm}$$

Since M is the midpoint of AB (perpendicular from centre bisects the chord):
$$AB = 2 \times AM = 2 \times 3 = 6 \text{ cm}$$

Note: Since x = 4 cm = OO′, point M coincides with O′. This means O′ lies on AB, confirming AB passes through O′.

$$\boxed{AB = 6 \text{ cm}}$$

Given: A circle with centre O. Two equal chords AB and CD intersect at point E inside the circle, i.e., AB = CD.

To Prove: AE = CE and BE = DE (corresponding segments are equal).

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Proof:

Since AB = CD and equal chords are equidistant from the centre:
$$OM = ON \quad \cdots (1)$$

Also, since perpendicular from centre bisects the chord:
$$MB = \frac{AB}{2} \quad \text{and} \quad ND = \frac{CD}{2}$$

Since AB = CD:
$$MB = ND \quad \cdots (2)$$

In △OME and △ONE:
$$OM = ON \quad \text{(from (1))}$$
$$\angle OME = \angle ONE = 90°$$
$$OE = OE \quad \text{(common)}$$

By RHS congruence:
$$\triangle OME \cong \triangle ONE$$

By CPCT:
$$ME = NE \quad \cdots (3)$$

Now:
$$BE = MB - ME = ND - NE = DE$$

Also:
$$AE = AB - BE = CD - DE = CE$$

Hence proved. AE = CE and BE = DE.

Given: A circle with centre O. Two equal chords AB and CD intersect at point E inside the circle. OE is joined.

To Prove: ∠OEM = ∠OEN, i.e., OE makes equal angles with the two chords (where M and N are feet of perpendiculars from O to AB and CD respectively).

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Proof:

Since AB = CD (equal chords of the same circle), they are equidistant from the centre:
$$OM = ON \quad \cdots (1)$$

In △OME and △ONE:
$$OM = ON \quad \text{(from (1))}$$
$$\angle OME = \angle ONE = 90°$$
$$OE = OE \quad \text{(common)}$$

By RHS congruence:
$$\triangle OME \cong \triangle ONE$$

By CPCT:
$$\angle OEM = \angle OEN$$

Hence proved. The line joining the point of intersection to the centre makes equal angles with the chords.

Given: Two concentric circles with common centre O. A line intersects the inner circle at B and C, and the outer circle at A and D.

To Prove: AB = CD

Construction: Draw OM perpendicular to the line AD from centre O.

Proof:

For the outer circle, OM ⊥ AD:
Since perpendicular from centre bisects the chord:
$$AM = MD \quad \cdots (1)$$

For the inner circle, OM ⊥ BC:
Since perpendicular from centre bisects the chord:
$$BM = MC \quad \cdots (2)$$

Subtracting (2) from (1):
$$AM - BM = MD - MC$$
$$AB = CD$$

Hence proved.

Given: A circle of radius 5 m. Let Reshma, Salma and Mandip be at points R, S and M on the circle. RS = SM = 6 m.

To Find: RM

Solution:

Let O be the centre of the circle. Draw OA ⊥ RS and OB ⊥ SM.

Since OA ⊥ RS and perpendicular from centre bisects the chord:
$$AS = \frac{RS}{2} = 3 \text{ m}$$

In △OAS:
$$OA^2 = OS^2 - AS^2 = 25 - 9 = 16$$
$$OA = 4 \text{ m}$$

Similarly, OB = 4 m (since SM = 6 m and radius = 5 m).

Let O be the origin. Place S at the top. Since RS = SM = 6 m and both chords are equal, by symmetry, R and M are symmetric about the line OS.

Let the perpendicular from O to RS meet RS at A, and from O to SM meet SM at B.

Using coordinate geometry: Let O = (0, 0).

Since OA ⊥ RS and OA = 4 m, place A along the y-axis direction.

Let S = (0, 5) (topmost point).

For chord RS: midpoint A is at distance 4 from O along OS direction.
$$A = (0, 4)$$

R and S are symmetric about A:
$$R = (-3, 4), \quad S = (3, 4)$$

Wait — let me redo with S as a general point.

Let S = (0, 5). Chord RS has length 6, midpoint A satisfies OA ⊥ RS.

Since RS is horizontal (by symmetry, if S is at top and R, M are symmetric):
$$A = (0, 4), \quad R = (-3, 4), \quad M = (3, 4)$$

Verify: $OR = \sqrt{9+16} = 5$ ✓, $OM = \sqrt{9+16} = 5$ ✓

Check SM: $SM = \sqrt{(3-0)^2+(4-5)^2} = \sqrt{9+1} = \sqrt{10} \neq 6$

Let me use a proper coordinate approach.

Let O = (0,0). Let S = (s_x, s_y) on the circle, so $s_x^2 + s_y^2 = 25$.

Let the perpendicular from O to RS have foot A. Since OA = 4 and AS = 3:

Place the coordinate system so that S = (0, 5).

For chord RS (length 6): The perpendicular from O to RS has length $\sqrt{25-9} = 4$.

Let RS be along a line. The foot of perpendicular from O to RS is at distance 4 from O.

Let A = foot on RS. Then $\vec{OA} \perp RS$ and $|OA| = 4$, $|AS| = 3$.

So $A = S - 3\hat{u}$ where $\hat{u}$ is unit vector along RS.

Also $\vec{OA} \cdot \hat{u} = 0$, meaning $\vec{OS} \cdot \hat{u} = 3$.

Let $\hat{u} = (\cos\theta, \sin\theta)$. Then $s_x\cos\theta + s_y\sin\theta = 3$.

Similarly for chord SM: $s_x\cos\phi + s_y\sin\phi = 3$ where $\hat{v} = (\cos\phi, \sin\phi)$ is direction of SM.

This is getting complex. Let me use a direct approach.

Let O = (0,0), S = (0, 5).

For chord RS of length 6: perpendicular from O to RS = 4.
Let R = $(x_1, y_1)$ on circle, $x_1^2+y_1^2=25$, $RS=6$.

Midpoint of RS = $\left(\frac{x_1}{2}, \frac{y_1+5}{2}\right)$, and this is perpendicular to RS from O.

Distance from O to midpoint = 4:
$$\frac{x_1^2}{4} + \frac{(y_1+5)^2}{4} = 16$$
$$x_1^2 + y_1^2 + 10y_1 + 25 = 64$$
$$25 + 10y_1 + 25 = 64$$
$$10y_1 = 14 \Rightarrow y_1 = 1.4$$

Then $x_1^2 = 25 - 1.96 = 23.04$, so $x_1 = \pm 4.8$.

By symmetry, $R = (-4.8, 1.4)$ and $M = (4.8, 1.4)$.

Verify SM: $SM = \sqrt{(4.8)^2+(1.4-5)^2} = \sqrt{23.04+12.96} = \sqrt{36} = 6$ ✓

Now:
$$RM = \sqrt{(4.8-(-4.8))^2 + 0^2} = \sqrt{(9.6)^2} = 9.6 \text{ m}$$

$$\boxed{RM = 9.6 \text{ m}}$$

Given: A circular park of radius 20 m. Three boys Ankur (A), Syed (S) and David (D) sit at equal distances on the boundary, forming an equilateral triangle inscribed in the circle of radius 20 m.

To Find: Side length of the equilateral triangle (length of string).

Solution:

Since the three boys are at equal distances, they form an equilateral triangle inscribed in the circle.

For an equilateral triangle inscribed in a circle of radius R, the relationship between side $a$ and circumradius R is:
$$R = \frac{a}{\sqrt{3}}$$

Derivation using perpendicular from centre:

Let O be the centre. Draw OM ⊥ AS where M is the midpoint of AS.

Let side of equilateral triangle = $a$, so $AM = \frac{a}{2}$.

In △OAS, O is the circumcentre. The centroid divides the median in ratio 2:1.

Median of equilateral triangle $= \frac{\sqrt{3}}{2}a$

Circumradius $= \frac{2}{3} \times \frac{\sqrt{3}}{2}a = \frac{a}{{\sqrt{3}}}$

So:
$$20 = \frac{a}{\sqrt{3}}$$
$$a = 20\sqrt{3} \text{ m}$$

$$\boxed{\text{Length of each string} = 20\sqrt{3} \approx 34.64 \text{ m}}$$

Given: ∠BOC = 30°, ∠AOB = 60°. D is a point on the circle on the arc other than arc ABC.

To Find: ∠ADC

Solution:

$$\angle AOC = \angle AOB + \angle BOC = 60° + 30° = 90°$$

By the theorem — the angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle:

$$\angle AOC = 2 \times \angle ADC$$

$$\angle ADC = \frac{\angle AOC}{2} = \frac{90°}{2} = 45°$$

$$\boxed{\angle ADC = 45°}$$

Given: Chord AB = radius of the circle (= r).

To Find: Angle subtended at a point on the minor arc and at a point on the major arc.

Solution:

Let O be the centre. Since OA = OB = AB = r, triangle OAB is equilateral.

$$\angle AOB = 60°$$

Angle at a point on the major arc:

Let P be a point on the major arc. By the inscribed angle theorem:
$$\angle APB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60° = 30°$$

Angle at a point on the minor arc:

Let Q be a point on the minor arc. AQBP is a cyclic quadrilateral (all on the circle). The sum of opposite angles:
$$\angle AQB + \angle APB = 180°$$
$$\angle AQB = 180° - 30° = 150°$$

$$\boxed{\text{Angle on major arc} = 30°, \quad \text{Angle on minor arc} = 150°}$$