Question

Two circles of equal radius of 2√3 intersect each other in such a way that both pass through
center of each other. The length of common chord is
(1) 8
(2) 6
(3)4
(4) 3​

Answer 1

Given :-

• Two circles of equal radius = 2√3 cm
• Both radius passes through center of each other.

To find :-

• Length of common chord.

Let :-

• O & O' be the center of both circle as given in above figure.
• OO' & O'O be the common radius.
• AB be the chord.

Construction :-

• Join AO, OB, BO' and O'A.

So, according to the question:-

If O and O' be the centres, then OO' will be the common radius.

∴OO′ = 2√3 cm

We know that radius of circle are always equal.

∴OA=OB=O′A=O′B

All Sides means radius are equal then it will be called as Rhombus.

∴ OAO'B is a rhombus.

We know that diagonals of a rhombus bisect each other at right angle.

∠AEO=90°

Value of OE is

= 2√3 / 2

= √3 cm

By pythagoras theorem:-

∴AO² = AE² +OE²

⟹ (2√3 )² = (AE)² + (√3)²

⟹ 2 × 3 = (AE)² + 3

⟹ 6 = (AE)² + 3

⟹ (AE)² = 6 - 3

⟹ (AE)² = 3

⟹ AE = √3 cm

The length of common chord is :-

⟹ 2 × AE

⟹ 2 × √3

⟹ 2√3

⟹ 3.47 ≈ 4 cm