🔑 Important Formulas, Rules & Key Points

• Perpendicular Bisector Rule: The perpendicular from the centre of a circle to a chord bisects the chord.
• Equidistant Chords: Equal chords of a circle (or of congruent circles) are equidistant from the centre.
• Common Chord Property: If two circles intersect at two points, then the line joining their centres is the perpendicular bisector of their common chord.
• Congruence Criteria: Standard Congruence rules like RHS, SSS, and SAS are frequently used to solve circle theorems.

Q1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Solution:

Let the two circles have centres O and O' with radii 5 cm and 3 cm respectively. The distance between their centres OO' = 4 cm.

Let the common chord be AB, intersecting OO' at point M. Since the line joining the centres is the perpendicular bisector of the common chord:

• AM = MB
• ∠AMO = ∠AMO' = 90°

Let OM = x. Therefore, O'M = 4 - x.

In right-angled triangle ΔAMO:
AM² = OA² - OM² (By Pythagoras Theorem)
AM² = 5² - x² = 25 - x² — (Equation 1)

In right-angled triangle ΔAMO':
AM² = O'A² - O'M²
AM² = 3² - (4 - x)² = 9 - (16 + x² - 8x) = -7 - x² + 8x — (Equation 2)

Equating Equation 1 and Equation 2:
25 - x² = -7 - x² + 8x
25 = -7 + 8x
8x = 32
x = 4 cm

Since x = 4 cm, the point M coincides with the centre O'. This means the common chord passes through the centre of the smaller circle.

Now, substitute x = 4 into Equation 1:
AM² = 25 - 4² = 25 - 16 = 9
AM = 3 cm

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

Answer: The length of the common chord is 6 cm.

Q2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:

Given: A circle with centre O has two equal chords AB and CD intersecting at point E inside the circle (i.e., AB = CD).

To Prove: AE = DE and CE = BE.

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

Proof:

• In right triangles ΔOME and ΔONE:
  1. ∠OME = ∠ONE = 90° (By construction)
  2. OE = OE (Common hypotenuse)
  3. OM = ON (Equal chords are equidistant from the centre)
  By RHS congruence rule, ΔOME ≅ ΔONE.
  Therefore, by CPCT: ME = NE — (Equation 1)
• Since OM ⊥ AB and ON ⊥ CD, they bisect the chords:
  AM = MB = ½ AB and DN = NC = ½ CD.
  Since AB = CD, we have: AM = DN and MB = NC — (Equation 2)
• To prove AE = DE, add Equation 1 and Equation 2:
  AM + ME = DN + NE ⇒ AE = DE (Proved)
• To prove CE = BE, subtract Equation 1 from MB = NC:
  NC - NE = MB - ME ⇒ CE = BE (Proved)

Q3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution:

Given: Chords AB = CD intersecting at E. O is the centre.

To Prove: ∠OEM = ∠OEN (where OM ⊥ AB and ON ⊥ CD).

Proof:

Construct OM ⊥ AB and ON ⊥ CD. In triangles ΔOME and ΔONE:

1. OM = ON (Equal chords are equidistant from the centre)
2. ∠OME = ∠ONE = 90°
3. OE = OE (Common side)

By RHS congruence criterion, ΔOME ≅ ΔONE.
Hence, by CPCT:
∠OEM = ∠OEN. (Proved)

Q4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.

Solution:

Construction: Draw OM ⊥ AD.

Proof:

For the outer circle, AD is a chord and OM ⊥ AD. Since the perpendicular from the centre bisects the chord:
AM = MD — (Equation 1)

For the inner circle, BC is a chord and OM ⊥ BC. Therefore:
BM = MC — (Equation 2)

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

Q5. Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandeep, Mandeep to Reshma. If the distance between Reshma and Salma and between Salma and Mandeep is 6m each, what is the distance between Reshma and Mandeep?

Solution:

Let R, S, and M represent Reshma, Salma, and Mandeep on the boundary of the circle with centre O. Radius OR = OS = OM = 5 m. Given RS = SM = 6 m.

Since RS = SM, the line joining the centre O to the vertex S acts as a perpendicular bisector of RM. Let OS intersect RM at P. Therefore, RP = PM and ∠RPO = 90°.

Let's find the area of triangle ΔORS using Heron's Formula:

• Sides of ΔORS: a = 5 m, b = 5 m, c = 6 m.
• Semi-perimeter s = (5 + 5 + 6) / 2 = 8 m.
• Area = √[s(s-a)(s-b)(s-c)] = √[8(8-5)(8-5)(8-6)] = √[8 × 3 × 3 × 2] = √144 = 12 m².

We can also compute the area of ΔORS using the base-height formula with OS as base and RP as altitude:
Area = ½ × Base × Height
12 = ½ × OS × RP
12 = ½ × 5 × RP
RP = 24 / 5 = 4.8 m

The total distance between Reshma and Mandeep is:
RM = 2 × RP = 2 × 4.8 = 9.6 m.

Answer: The distance between Reshma and Mandeep is 9.6 m.

Q6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

Solution:

Let A, S, and D represent Ankur, Syed, and David respectively. Since they are sitting at equal distances, ΔASD is an equilateral triangle. Let the side length of ΔASD be 2x meters.

Let the radius of the circle be R = 20 m. The altitude of an equilateral triangle with side 2x is:
h = (√3 / 2) × Side = (√3 / 2) × 2x = √3x

Since the triangle is equilateral, its circumcentre coincides with its centroid O. The centroid divides the altitude in the ratio 2:1.
Therefore, the distance AO (circumradius) is (2/3) × h:
AO = (2 / 3) × √3x = (2 / √3)x

We are given circumradius AO = 20 m:
20 = (2 / √3)x
x = 10√3 m

The side length of the equilateral triangle (length of each string) is:
2x = 2 × 10√3 = 20√3 m

Answer: The length of the string of each phone is 20√3 m (or approx 34.64 m).

🚀 Aptitude Corner (NTSE / Olympiad Focus)

The Intersecting Chords Theorem: If two chords of a circle, AB and CD, intersect at a point P (inside or outside the circle), then:
PA × PB = PC × PD

Practice Question: Two chords AB and CD of a circle intersect inside the circle at point P. If PA = 4 cm, PB = 6 cm, and PC = 3 cm, find the length of PD.

Quick Calculation:
PA × PB = PC × PD
4 × 6 = 3 × PD
24 = 3 × PD ⇒ PD = 8 cm.