Circles - NCERT Solutions

Solution:

Let the circles with centers O and O' have radii 5 cm and 3 cm respectively.

Distance between centers OO' = 4 cm

Let AB be the common chord and M be its midpoint.

Then OM ⊥ AB and O'M ⊥ AB.

Let OM = x cm, then O'M = (4 - x) cm

In right triangle OMA:

AM² = OA² - OM² = 25 - x²

In right triangle O'MA:

AM² = O'A² - O'M² = 9 - (4 - x)²

Equating both expressions for AM²:

25 - x² = 9 - (16 - 8x + x²)

25 - x² = 9 - 16 + 8x - x²

25 = -7 + 8x

8x = 32 ⇒ x = 4

Then AM² = 25 - 16 = 9 ⇒ AM = 3 cm

Therefore, length of common chord AB = 2 × AM = 2 × 3 = 6 cm

Solution:

Given: A circle with center O. Two equal chords AB and CD intersect at point E.

To prove: AE = CE and BE = DE

Construction: Draw OM ⊥ AB and ON ⊥ CD. Join OE.

Proof:

Since AB = CD (Given)

∴ OM = ON (Equal chords are equidistant from the center)

In right triangles OME and ONE:

• OM = ON (Proved)
• OE = OE (Common)

∴ ΔOME ≅ ΔONE (RHS congruence rule)

∴ ME = NE (CPCT)

Now, AM = MB (Perpendicular from center bisects the chord)

Similarly, CN = ND

Since AB = CD, we have AM + MB = CN + ND

⇒ 2AM = 2CN ⇒ AM = CN

Now, AE = AM - ME and CE = CN - NE

Since AM = CN and ME = NE, we get AE = CE

Similarly, BE = DE

Hence, the segments of one chord are equal to corresponding segments of the other chord.

Solution:

Given: A circle with center O. Two equal chords AB and CD intersect at point E.

To prove: ∠OEA = ∠OED

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Proof:

Since AB = CD (Given)

∴ OM = ON (Equal chords are equidistant from the center)

In right triangles OME and ONE:

• OM = ON (Proved)
• OE = OE (Common)

∴ ΔOME ≅ ΔONE (RHS congruence rule)

∴ ∠OEM = ∠OEN (CPCT)

But ∠OEA = ∠OEM and ∠OED = ∠OEN

∴ ∠OEA = ∠OED

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

Solution:

Given: Two concentric circles with center O. A line intersects the circles at A, B, C, D (in order).

To prove: AB = CD

Construction: Draw OM ⊥ AD.

Proof:

Since OM ⊥ AD,

∴ AM = MD (Perpendicular from center bisects the chord)

Similarly, BM = MC

Now, AB = AM - BM

And CD = MD - MC

Since AM = MD and BM = MC, we get AB = CD

Hence proved.

Solution:

Let R, S, M represent positions of Reshma, Salma and Mandip respectively.

Given: OR = OS = OM = 5 m (radius)

RS = SM = 6 m

To find: RM

Let's find ∠ROS:

In ΔORS, by cosine rule:

cos(∠ROS) = (OR² + OS² - RS²) / (2 × OR × OS)

= (25 + 25 - 36) / (2 × 5 × 5) = 14/50 = 7/25

Similarly, in ΔOSM:

cos(∠SOM) = (OS² + OM² - SM²) / (2 × OS × OM)

= (25 + 25 - 36) / (2 × 5 × 5) = 14/50 = 7/25

∴ ∠ROS = ∠SOM

Now, ∠ROM = ∠ROS + ∠SOM = 2∠ROS

In ΔORM, by cosine rule:

RM² = OR² + OM² - 2 × OR × OM × cos(∠ROM)

= 25 + 25 - 2 × 5 × 5 × cos(2∠ROS)

= 50 - 50 × [2cos²(∠ROS) - 1] (Using cos2θ = 2cos²θ - 1)

= 50 - 50 × [2 × (7/25)² - 1]

= 50 - 50 × [2 × 49/625 - 1]

= 50 - 50 × [98/625 - 625/625]

= 50 - 50 × (-527/625)

= 50 + 26350/625

= 50 + 42.16 = 92.16

∴ RM = √92.16 ≈ 9.6 m

Alternatively, we can use the property that in a circle, equal chords subtend equal angles at the center.

Since RS = SM = 6 m, they subtend equal angles at the center.

∴ ∠ROS = ∠SOM

Also, R, S, M are three points on the circle, so the chord RM subtends an angle of 2∠ROS at the center.

Using the formula: chord length = 2 × radius × sin(angle at center/2)

First, find ∠ROS:

RS = 2 × 5 × sin(∠ROS/2) = 6

⇒ sin(∠ROS/2) = 6/10 = 3/5

Then RM = 2 × 5 × sin(∠ROM/2) = 10 × sin(∠ROS) (since ∠ROM = 2∠ROS)

Now, sin(∠ROS) = 2 × sin(∠ROS/2) × cos(∠ROS/2) = 2 × (3/5) × (4/5) = 24/25

∴ RM = 10 × 24/25 = 240/25 = 9.6 m

Hence, the distance between Reshma and Mandip is 9.6 m.

Solution:

The three boys are sitting at equal distances on the boundary of a circle of radius 20 m.

This means they form an equilateral triangle inscribed in the circle.

For an equilateral triangle inscribed in a circle:

Side length = √3 × radius

Proof: In an equilateral triangle inscribed in a circle, the radius R and side a are related by:

a = √3 × R

This can be derived from the fact that the circumradius of an equilateral triangle is a/√3.

So, length of string = side of equilateral triangle = √3 × 20 = 20√3 m

Alternatively, using chord length formula:

Chord length = 2R sin(θ/2), where θ is the angle subtended at center.

For three equally spaced points, θ = 120°

∴ Chord length = 2 × 20 × sin(60°) = 40 × (√3/2) = 20√3 m

Hence, the length of the string of each phone is 20√3 m.

Solution:

Given: ∠AOB = 60°, ∠BOC = 30°

Then ∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°

Now, ∠ADC is the angle subtended by arc AC at point D on the remaining part of the circle.

By theorem: The angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOC = 2∠ADC

⇒ 90° = 2∠ADC

⇒ ∠ADC = 45°

Hence, ∠ADC = 45°

Solution:

Let the circle have center O and radius r.

Given: Chord AB = r

In ΔOAB:

OA = OB = r (radii)

AB = r (given)

∴ ΔOAB is equilateral

⇒ ∠AOB = 60°

Now, the angle subtended by chord AB at any point on the minor arc is half the angle subtended at the center by the major arc.

Major arc AB subtends reflex ∠AOB = 360° - 60° = 300° at the center.

∴ Angle subtended by chord AB at a point on the minor arc = 1/2 × 300° = 150°

Angle subtended by chord AB at a point on the major arc = 1/2 × 60° = 30°

Hence, the angle subtended by the chord is 150° on the minor arc and 30° on the major arc.

Solution:

Given: ∠PQR = 100°

We need to find ∠OPR.

Consider the arc PR.

∠PQR is the angle subtended by arc PR at point Q on the remaining part of the circle.

By theorem: The angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle.

∴ ∠POR = 2∠PQR = 2 × 100° = 200°

Now, in ΔOPR:

OP = OR (radii)

∴ ∠OPR = ∠ORP (angles opposite equal sides)

Let ∠OPR = ∠ORP = x

Then, in ΔOPR:

∠OPR + ∠ORP + ∠POR = 180°

⇒ x + x + 200° = 180°

This gives 2x = -20°, which is not possible.

We need to consider that ∠POR = 200° is the reflex angle.

The actual angle in triangle OPR is 360° - 200° = 160°

So, x + x + 160° = 180°

⇒ 2x = 20°

⇒ x = 10°

Hence, ∠OPR = 10°

Solution:

Given: ∠ABC = 69°, ∠ACB = 31°

In ΔABC:

∠BAC = 180° - (69° + 31°) = 180° - 100° = 80°

Now, points A, B, C, D are concyclic (lie on the same circle).

In a cyclic quadrilateral, angles in the same segment are equal.

∠BDC and ∠BAC are angles in the same segment (segment BC).

∴ ∠BDC = ∠BAC = 80°

Hence, ∠BDC = 80°

Solution:

Given: ∠BEC = 130°, ∠ECD = 20°

In ΔBEC:

∠EBC = 180° - (130° + 20°) = 30°

Now, ∠BAC and ∠BDC are angles in the same segment (segment BC).

But ∠BDC = ∠EDC = 180° - ∠ECD - ∠CED

In ΔECD:

∠CED = 180° - (20° + ∠CDE)

We can use the property of vertically opposite angles:

∠BEC and ∠AED are vertically opposite, so ∠AED = 130°

In ΔAED:

∠EAD = 180° - (130° + ∠ADE)

Alternatively, we can use the property:

∠BAC and ∠BDC are angles in the same segment.

∠BDC = ∠BEC - ∠ECD (exterior angle property in ΔBEC)

Wait, let's think differently:

In ΔBEC:

∠EBC = 180° - (130° + 20°) = 30°

Now, ∠BDC and ∠BAC are angles in the same segment (segment BC).

But ∠BDC = ∠BEC - ∠ECD = 130° - 20° = 110°

This is not correct as angles in triangle should sum to 180°.

Let's use the property: Angles in the same segment are equal.

∠BAC and ∠BDC are in the same segment BC.

We need to find ∠BDC.

In ΔECD:

∠CED = 180° - 130° = 50° (linear pair with ∠BEC)

Then ∠CDE = 180° - (20° + 50°) = 110°

So ∠BDC = ∠CDE = 110°

Therefore, ∠BAC = ∠BDC = 110°

But this seems too large. Let's verify:

In cyclic quadrilateral ABDC:

∠BAC + ∠BDC = 180°

If ∠BAC = 110°, then ∠BDC = 70°

This seems more reasonable.

Let me recalculate:

In ΔECD:

∠CED = 180° - 130° = 50°

∠ECD = 20°

∴ ∠CDE = 180° - (50° + 20°) = 110°

Now, in cyclic quadrilateral ABDC:

∠BAC + ∠BDC = 180°

But ∠BDC = ∠CDE = 110°

Then ∠BAC = 180° - 110° = 70°

Hence, ∠BAC = 70°

Solution:

Given: ∠DBC = 70°, ∠BAC = 30°

First, find ∠BCD:

∠BDC and ∠BAC are angles in the same segment (segment BC).

∴ ∠BDC = ∠BAC = 30°

In ΔBCD:

∠BCD = 180° - (∠DBC + ∠BDC) = 180° - (70° + 30°) = 80°

Now, if AB = BC:

In ΔABC, AB = BC, so it's isosceles.

∴ ∠BCA = ∠BAC = 30°

Now, ∠BCD = ∠BCA + ∠ACD

⇒ 80° = 30° + ∠ACD

⇒ ∠ACD = 50°

Now, ∠ECD = ∠ACD = 50° (since E is on AC)

Hence, ∠BCD = 80° and ∠ECD = 50°

Solution:

Given: ABCD is a cyclic quadrilateral with AC and BD as diameters.

To prove: ABCD is a rectangle.

Proof:

Since AC is a diameter, ∠ABC = 90° (angle in a semicircle)

Similarly, since AC is a diameter, ∠ADC = 90°

Since BD is a diameter, ∠BAD = 90°

Similarly, since BD is a diameter, ∠BCD = 90°

So all angles of quadrilateral ABCD are 90°.

Hence, ABCD is a rectangle.

Solution:

Given: A trapezium ABCD with AB || CD and AD = BC.

To prove: ABCD is cyclic.

Construction: Draw DE ⊥ AB and CF ⊥ AB.

Proof:

In right triangles AED and BFC:

• AD = BC (Given)
• DE = CF (Distance between parallel lines)

∴ ΔAED ≅ ΔBFC (RHS congruence rule)

∴ ∠DAE = ∠CBF (CPCT)

⇒ ∠DAB = ∠CBA

Now, since AB || CD, we have ∠DAB + ∠ADC = 180° (co-interior angles)

But ∠DAB = ∠CBA

∴ ∠CBA + ∠ADC = 180°

This shows that the sum of opposite angles of quadrilateral ABCD is 180°.

Hence, ABCD is cyclic.

Solution:

Given: Two circles intersecting at B and C.

Line ABD intersects the circles at A, B, D.

Line PBQ intersects the circles at P, B, Q.

To prove: ∠ACP = ∠QCD

Proof:

In the first circle, ∠ACP and ∠ABP are angles in the same segment (segment AP).

∴ ∠ACP = ∠ABP ...(1)

In the second circle, ∠QCD and ∠QBD are angles in the same segment (segment QD).

∴ ∠QCD = ∠QBD ...(2)

But ∠ABP and ∠QBD are vertically opposite angles.

∴ ∠ABP = ∠QBD ...(3)

From (1), (2) and (3):

∠ACP = ∠QCD

Hence proved.

Solution:

Given: ΔABC

Circles with diameters AB and AC intersect at point D.

To prove: D lies on BC.

Proof:

Since AB is diameter of first circle, ∠ADB = 90° (angle in a semicircle)

Since AC is diameter of second circle, ∠ADC = 90° (angle in a semicircle)

Now, ∠ADB + ∠ADC = 90° + 90° = 180°

This means points B, D, C are collinear (since the angles form a linear pair).

Hence, D lies on BC.

Solution:

Given: ΔABC and ΔADC are right triangles with common hypotenuse AC.

To prove: ∠CAD = ∠CBD

Proof:

Since ΔABC is right-angled, ∠ABC = 90°

Since ΔADC is right-angled, ∠ADC = 90°

So, in quadrilateral ABCD,

∠ABC + ∠ADC = 90° + 90° = 180°

This means ABCD is cyclic (if sum of opposite angles is 180°, quadrilateral is cyclic).

Now, in cyclic quadrilateral ABCD,

∠CAD and ∠CBD are angles in the same segment (segment CD).

∴ ∠CAD = ∠CBD

Hence proved.

Solution:

Given: ABCD is a cyclic parallelogram.

To prove: ABCD is a rectangle.

Proof:

Since ABCD is a parallelogram, opposite angles are equal.

So, ∠A = ∠C and ∠B = ∠D

Since ABCD is cyclic, sum of opposite angles is 180°.

So, ∠A + ∠C = 180°

But ∠A = ∠C, so 2∠A = 180° ⇒ ∠A = 90°

Similarly, ∠B = 90°

So all angles of parallelogram ABCD are 90°.

Hence, ABCD is a rectangle.

Solution:

Given: AB and CD are two chords intersecting at E.

PQ is a diameter through E such that ∠AEQ = ∠DEQ.

To prove: AB = CD

Construction: Draw OL ⊥ AB and OM ⊥ CD.

Proof:

In ΔOLE and ΔOME:

• ∠LEO = ∠MEO (Given, since ∠AEQ = ∠DEQ)
• ∠OLE = ∠OME = 90° (Construction)
• OE = OE (Common)

∴ ΔOLE ≅ ΔOME (AAS congruence rule)

∴ OL = OM (CPCT)

Now, chords equidistant from the center are equal.

∴ AB = CD

Hence proved.

Solution:

Given: AB is diameter, CD = radius.

To prove: ∠AEB = 60°

Construction: Join OC, OD and BC.

Proof:

In ΔOCD:

OC = OD = CD (All equal to radius)

∴ ΔOCD is equilateral

⇒ ∠COD = 60°

Now, ∠CBD = 1/2 ∠COD (Angle at center is twice angle at circumference)

= 1/2 × 60° = 30°

Also, ∠ACB = 90° (Angle in a semicircle)

In ΔBCE:

∠BCE = 180° - ∠ACB = 180° - 90° = 90°

∴ ∠CEB = 180° - (∠BCE + ∠CBE) = 180° - (90° + 30°) = 60°

But ∠CEB = ∠AEB (Same angle)

∴ ∠AEB = 60°

Hence proved.

Solution:

Given: ∠DBC = 55°, ∠BAC = 45°

To find: ∠BCD

∠CAD = ∠DBC = 55° (Angles in the same segment)

∴ ∠DAB = ∠CAD + ∠BAC = 55° + 45° = 100°

Now, in cyclic quadrilateral ABCD:

∠DAB + ∠BCD = 180° (Sum of opposite angles)

⇒ 100° + ∠BCD = 180°

⇒ ∠BCD = 80°

Hence, ∠BCD = 80°

Solution:

Given: Two circles intersect at A and B.

AD is diameter of first circle, AC is diameter of second circle.

To prove: B lies on DC.

Construction: Join AB.

Proof:

Since AD is diameter, ∠ABD = 90° (Angle in a semicircle)

Since AC is diameter, ∠ABC = 90° (Angle in a semicircle)

∴ ∠ABD + ∠ABC = 90° + 90° = 180°

This means D, B, C are collinear.

Hence, B lies on DC.

Solution:

Given: ABCD is a quadrilateral.

AH, BE, CF, DH are internal angle bisectors of ∠A, ∠B, ∠C, ∠D respectively.

They form quadrilateral EFGH.

To prove: EFGH is cyclic.

Proof:

In ΔAEB:

∠AEB = 180° - (∠EAB + ∠EBA) = 180° - 1/2(∠A + ∠B)

Similarly, in ΔCGD:

∠CGD = 180° - (∠GCD + ∠GDC) = 180° - 1/2(∠C + ∠D)

Now, ∠FEH = ∠AEB (Vertically opposite angles)

And ∠FGH = ∠CGD (Vertically opposite angles)

∴ ∠FEH + ∠FGH = 180° - 1/2(∠A + ∠B) + 180° - 1/2(∠C + ∠D)

= 360° - 1/2(∠A + ∠B + ∠C + ∠D)

= 360° - 1/2 × 360° = 360° - 180° = 180°

So, in quadrilateral EFGH, sum of opposite angles is 180°.

Hence, EFGH is cyclic.