Let P be the point on the circlex2+y2=9 , Q a point on the line 7x+y+3=0 and the perpendicular bisector of PQ be the line x−y+1=0 . Then the coordinate of P are

Any point on the line 7x+y+3=0 is Q(t,−3−7t) , t∈R Now P(h, k) is image of point Q in the line x−y+1=0 Then h−t1=k−(−3−7t)−1                       =2(t−(−3−7t)+1)1+1                        =−8t−4 ⇒(h,k)=(−7t−4,t+1) This point lies on the circlex2+y2=9 . So (−7t−4)2+(t+1)2=9 ⇒50t2+58t+8=0 ⇒25t2+29t+4=0 (25t+4)(t+1)=0 ⇒t=−4/25,t=−1 ⇒(h,k)≡(−7225,2125)or(3,0)