Let | z -i 2 z +i |= 1 3 , z C , be the equation of a circle with center at C . If the area of the triangle, whose vertices are at the points (0,0), C and ( , 0) is 11 square units, then ^2 equals:

Step-by-step solution

Let z=x+i y z =x-i y aligned & 3| z -i|=1|2 z +i| & =3 (x-(y+1) i|=|2 x+i(1-2 y)| & =3 x^2+(y+1)^2 = (2 x)^2+(1-2 y)^2 & =9 (x^2+y^2+2 y+1 )=4 x^2+4 y^2-4 y+1 & 5 x^2+5 y^2+22 y+8=0 & Centre (0,- 11 5 ) aligned Area of = 1 2 | | | -11 5 |=11 aligned & | |=10 & ^2=100 aligned