2 circles of radii 10cm and 6cm intersects at 2 points and the distance between their centers is 8cm. 

Find the length of the common chord . 

Let  X and Y be the centres of circles having radius 10 cm and 6 cm respectively. Suppose these circles intersect in A and B. XY intersect AB in P.

In ΔXAY and ΔXBY,

AX = BX      (Radius of the circle having centre X)

XY = XY      (Common)

AY = BY      (Radius of the circle having centre Y)

∴ ΔXAY ΔXBY    (SSS congruence criterion)

⇒ ∠AXY = ∠BXY  (CPCT)

In ΔXAP and ΔXBP,

AX = BX    (Radius of the circle)

∠AXP = ∠BXP    (∠AXY = ∠BXY)

XP = XP    (Common)

∴ ΔXAP ΔXBP    (SAS congruence axiom)

⇒ ∠APX = ∠BPX  (CPCT)

∠APX + ∠BPX = 180°    (Linear pair)  

2∠APX = 180°    (∠APX = ∠BPX)

⇒ ∠APX = 90°  

⇒ XP ⊥ AB

⇒ P is the mid point AB  (Perpendicular from the centre to the chord, bisect the chord)

⇒ AP = PB

Let XP = x,

∴ PY = 8 – x

In right ΔAPX,

AX² = XP² + AP²

∴ (10)² = x² + AP²

⇒ AP² = (100  – x²)    ...(1)

In right ΔAPY,

AY² = YP² + AP²

∴ (6)² = (8 – x)²  + AP²

⇒ AP² = 36 – (8 – x)    ...(2)

Form (1) and (2), we have

100 – x² = 36 – (8 – x)²

∴ 100 – x² = 36 – (64 + x² – 16x)

⇒ 100 = 16x – 28

⇒ 16x = 128

⇒ x = 8

From (1), we have

AP² = 100 – 64 = 36      (x = 8)

⇒ AP = 6 cm

AB = AP + PB = 2 AP = 2 × 6 cm = 12 cm     (AP = PB)

Thus, the length of the common chord is 12 cm.