Solve this : A v a r i a b l e c i r c l e C h a s t h e e q u a t i o n x 2 + y 2 - 2 ( t 2 - 3 t + 1 ) x - 2 ( t 2 + 2 t ) y + t = 0 , w h e r e t i s a p a r a m e t e r . I f t h e p o w e r o f p o i n t P ( a , b ) w . r . t t h e c i r c l e C i s c o n s tan t t h e n t h e o r f e r e d p a i r ( a , b ) i s ( A ) 1 10 , - 1 10 ( B ) - 1 10 , 1 10 ( C ) 1 10 , 1 10 ( D ) - 1 10 , - 1 10 Share with your friends Share 1 Aarushi Mishra answered this C:x2+y2-2t2-3t+1x-2t2+2t+1y+t=0C:x2+y2-2xt2+6xt-2x-2yt2-4yt-2y+t=0C:x2+y2-2x-2y-2xt2-2yt2+6xt-4yt+t=0C:x2+y2-2x-2y-2t2x+y+t6x-4y+1=0Power of a point w.r.t. point Px1,y1 is defined as the value obtained by putting P in the given equation, here power of point Pa,b w.r.t. C=a2+b2-2a-2b-2t2a+b+t6a-4b+1Since Power is independent of t, therefore coefficients ot t and t2 must be 0 a+b=0_______________(1)6a-4b+1=0__________________(2)from equation 1b=-aput in equation 26a+4a+1=010a=-1a=-110b=110 Option B 1 View Full Answer