Q6 6 . F r o m t h e p o i n t A 0 , 3 i n t h e c i r c l e x 2 + 4 x + y - 3 2 = 0 a c h o r d A B i s d r a w n & e x t e n d e d t o a p o i n t M s u c h t h a t A M = 2 A B . T h e e q u a t i o n o f t h e l o c u s o f M i s : a x 2 + 8 x + y 2 = 0 b x 2 + 8 x + y - 3 2 = 0 c x - 3 2 + 8 x + y 2 = 0 d x 2 + 8 x + 8 y 2 = 0 Share with your friends Share 1 Neha Sethi answered this Dear student Given:x2+4x+y-32=0⇒x2+4x+y-32+22-22=0⇒x+22+y-32=4Let the coordinates be Mh,k ,where B is a mid point of A and M.Bh2,k+32But AB is the chord of circle x2+4x+y-32=0Thus, B must satisfy above equation.So, h24+4h2+12k+3-32=0⇒h24+4h2+k+324+9-3k+3=0⇒h2+8h+k+32+36-12k+34=0⇒h2+8h+k2+9+6k+36-12k-36=0⇒h2+k2+8h-6k+9=0So locus of M is the circlex2+y2+8x-6y+9=0 ie,x2+8x+y2-6y+9=0x2+8x+y-32=9-9x2+8x+y-32=0 is the required answer Regards 0 View Full Answer