If the locus of z C , such that Re ( z-1 2 z+ i )+ Re ( z -1 2 z - i )=2 is a circle of radius r and center (a, b) then 15 a b r^2 is equal to :

Step-by-step solution

aligned & Re ( z-1 2 z+i )+ Re ( z -1 2 z -i )=2 & Here, z-1 2 z+i = ( z -1 2 z -i )=2 & = Re ( z-1 2 z+i )+ Re ( z -1 2 z+i )=2 & =2 Re ( z-1 2 z+1 )=2 Re ( z-1 2 z+i )=1 aligned Let z=x+i y aligned & Re ( (x-1)+i y 2 x+i(2 y+1) )=1 Re [ ((x-1)+i y)(2 x-i(y+1) (2 x+i(2 y+1)(2 x-i(2 y+1)) ]=1 & 2 x(x-1)+y(2 y+1) 4 x^2+(2 y+1)^2 =1 & 2 x^2-2 x+2 y^2+y=4 x^2+4 y^2+1+4 y & 2 x^2+2 y^2+3 y+2 x+1=0 & x^2+y^2+x+ 3 2 y+ 1 2 =0 & centre = ( -1 2 , -3 4 ), r= 1 4 + 9 16 - 1 2 = 5 4 & a= -1 2 , b= -3 4 , r^2= 5 16 & 15 a b r