the common chord be AB and P \and\ Q be the centers of the two circles. AP=5cm \and\ AQ=3cm. PQ=4cm ...given Now, segPQ⊥chord AB AR=RB=1/2AB....perpendicular from center to the chord, bisects the chord Let PR=x cm so RQ=(4−x)cm In/_\ARP,AP^2 =AR^2+PR^2 AR^2=5^2−x^2...(1) In /_\ARQ,AQ^2=AR^2+QR^2 AR^2=3^2−(4−x)^2...(2) 5^2−x^2=3^2−(4−x)^2...from (1) & (2) 25−x^2=9−(16−8x+x^2) 25−x^2=−7+8x−x^2 32=8x x=4 substituting in equation(1) we get , AR^2=25-16=9 AR=3cm AB=2xxAR=2xx3=6 AB=6cmSo length of common chord is6 cm.
