🔑 Important Formulas, Rules & Key Points

• Circle: The collection of all points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle.
• Congruent Circles: Two circles are congruent if and only if they have equal radii (i.e., if you superimpose one on the other, they cover each other completely).
• Theorem 9.1: Equal chords of a circle (or of congruent circles) subtend equal angles at their centres.
• Theorem 9.2 (Converse of Theorem 9.1): If the angles subtended by the chords of a circle (or of congruent circles) at the centre are equal, then the chords are equal.

Question 1: Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Proof:

Let us consider two congruent circles C₁ and C₂ with centres O and O' respectively.

Let AB be a chord of circle C₁ and CD be a chord of circle C₂, such that AB = CD.

We need to prove that: ∠AOB = ∠CO'D.

Now, in triangles ΔAOB and ΔCO'D:

1. OA = O'C (Radii of congruent circles are equal)
2. OB = O'D (Radii of congruent circles are equal)
3. AB = CD (Given)

Therefore, by SSS (Side-Side-Side) Congruence Criterion:

ΔAOB ≅ ΔCO'D

By CPCT (Corresponding Parts of Congruent Triangles):

∠AOB = ∠CO'D

Hence, proved!

Question 2: Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Proof:

Let us consider two congruent circles C₁ and C₂ with centres O and O' respectively.

Let AB be a chord of circle C₁ and CD be a chord of circle C₂, such that the subtended angles at the centres are equal, i.e., ∠AOB = ∠CO'D.

We need to prove that: AB = CD.

Now, in triangles ΔAOB and ΔCO'D:

1. OA = O'C (Radii of congruent circles are equal)
2. ∠AOB = ∠CO'D (Given)
3. OB = O'D (Radii of congruent circles are equal)

Therefore, by SAS (Side-Angle-Side) Congruence Criterion:

ΔAOB ≅ ΔCO'D

By CPCT (Corresponding Parts of Congruent Triangles):

AB = CD

Hence, proved!

🚀 Aptitude Corner (For NTSE, Olympiads & Competitive Exams)

Practice Question:

Two concentric circles have radii of 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Quick Concept & Solution:

Let the common centre be O. The chord AB of the larger circle touches the smaller circle at point P. This makes OP a radius of the smaller circle, and AB is tangent to it at P. Thus, OP ⊥ AB.

In right-angled triangle ΔOPA (where OA is the radius of the larger circle = 5 cm, and OP = 3 cm):

OA² = OP² + AP²
5² = 3² + AP²
25 = 9 + AP²
AP² = 16 ⇒ AP = 4 cm

Since the perpendicular from the centre bisects the chord, the total length of the chord is:

AB = 2 × AP = 2 × 4 cm = 8 cm.