Given : Circle C (P, 3) and circle C (Q, 5) are intersecting at points A and B.
Construction: Join PA and QA. Draw PM as bisector of chord AB.
Proof: AB is chord of circle C (P, 3) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the

chord.]
Hence, ∠PMA = 90°
Let, PM = X, therefore, QM = 4 – x
In ΔAPM, using Pythagoras theorem
AM² = AP² – PM² …(1)
And in ΔAPM, using Pythagoras theorem
AM² = AQ² – QM² …(2)
From the equation (1) and (2), we get
AP² – PM² = AQ² – QM²
⇒ 3² – x² = 5² -(4 – x)² ⇒ 9 – x² = 25 – (16 + x² – 8x)
⇒ 9 – 9 = 8x ⇒ x = 0/8 = 0
from the equation (1), AM² = 3² – 0² = 9 ⇒ AM = 3
⇒ AB = 2AM = 6