Let P be a point on the circle x2+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 coordinates of P are

Let the coordinates of P and Q be (h, k) and x1,y1 respectively. Then,        h2+k2=9                …(i)and 7x1+y1+3=0        …(ii)Since PQ is perpendicular to the line x−y+1=0.∴ y1−kx1−h=−1⇒x1+y1=h+k         …(iii)The mid point of PQ lies on x-y+1 = 0∴ x1+h2−y1+k2+1=0⇒x1−y1=−(h−k+2)   …(iv)From (iii) and (iv), we get x1=k−1,y1=h+1Substituting x1 , y1 in (ii), we geth+7k=3⇒h=3−7kPutting the value of h in h2+k2=9, we get50k2−42k=0⇒k=0,21/25.Hence, the coordinates of P are (3,0) or, (−72/25,21/25)