Question

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.

Answer 1

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm 

O'A = O'B = 3 cm 

OO' will be the perpendicular bisector of chord AB. 

∴ AC = CB 

It is given that, OO' = 4 cm 

Let OC be x. 

Therefore, O'C will be 4 − x. 

In ∆OAC, 

OA² = AC² + OC²

⇒ 5 ² = AC² + x² 

⇒ 25 − x² = AC² ... (1) 

In ∆O'AC, 

O'A² = AC² + O'C² 

⇒ 3 ² = AC² + (4 − x)² 

⇒ 9 = AC² + 16 + x² − 8x 

⇒ AC² = − x² − 7 + 8x ... (2) 

From equations (1) and (2), we obtain 

25 − x² = − x² − 7 + 8x 

8x = 32 x = 4 

Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle. 

AC² = 25 − x² = 25 − 4² = 25 − 16 = 9 

∴AC = 3 m 

Length of the common chord AB = 2 AC = (2 × 3) m = 6m.

Answer 2

Let's assume the following :
AB be the common chord
P and Q be the centers of the two circles.
∴ AP = 5 cm and AQ = 3 cm.
Given :
PQ = 4 cm
Now, seg PQ ⊥ chord AB
∴ AR = RB = \(\frac{1}{2}AB\)  … perpendicular from center to the chord, bisects the chord
Let's suppose :
PR = x cm
So, RQ will be (4 - x)cm
Now, in △ARP,
AQ² = AR² + PR²
AR² = 5² - x² …. (i)
Now, in △ARQ,
AQ² = AR² + QR²
AR² = 3² - (4 - x)² …… (ii)
∴5² - x² = 3² - (4 - x)² …. from eqn (i) and (ii)
25 - x² = 9 - (16 - 8x + x²)
25 - x² = -7 + 8x - x²
32 = 8x
Therefore, x = 4
Substitute x = 4 in equation (i), we get
AR² = 25 - 16 = 9
Therefore, AR = 3 cm.
AB = 2 × AR 
= 2 × 3
So, AB = 6 cm.
So, length of common chord AB is 6 cm.