A line is drawn through a fixed point P (α, β) to cut the circle  x2+y2=r2  at A and  B .Then PA⋅PB is equal to

The equation of any line throughP (α, β) is x−αcos⁡θ=y−βsin⁡θ=k(say)Any point on this line is  (α+ k cos θ, β + k sinθ). This point lies on the given circle if(α+kcos⁡θ)2+(β+ksin⁡θ)2=r2⇒k2+2k(αcos⁡θ+βsin⁡θ)+α2+β2−r2=0             (i)This equation, being quadratic in k, gives two values of k and hence the distances of two points A and B on the circle from the point P. Let PA=k1,PB=k2, where k1,k2 are the roots of equation (i)Then,  PAPB =k1k2=α2+β2−r2ALITER  PAPB is the power of the point P(α,β) ) with respect to the circle x2+y2=r2 . Therefore , PAPB=α2+β2−r2