If θ  is the angle of intersection of two circlex2+y2=a2  and (x−c)2+y2=b2,  then the length of common chord of two circle is :

We have the two circles x2+y2=a2,  (x−c)2+y2=b2  radius of first circle=aLet ∠OPM=α, ∠CPM=β,∴  ∠OPC=α+β=θLet PQ=d;  radius of second circle=bcosα=PMa=d2a, cosβ=d2bNow cosθ=cos(α+β) =cosαcosβ−sinαsinβsquaring on both sides, we get sin2αsin2β =cos2αcos2β+cos2θ−2cosθcosαcosβ                      = 1−cos2α−cos2β+cos2αcos2β                     =cos2αcos2β+cos2θ−2cosθcosαcosβ⇒sin2θ=cos2α+cos2β−2cosθcosαcosβ∴  sin2θ=d24a2+d24b2−2d24abcosθ⇒  4a2b2sin2θ=d2(b2+a2−2abcosθ)∴  d=2absinθa2+b2−2abcosθ