Let a=1+i and z=x+i y . If the curve z z +a z+ a z -4=0 is cut by the straight line (z+ z )-i(z- z )+2=0 at two points A and B , then the equation of the circle passing through the origin, A and B is

Step-by-step solution

Circle z z +a z+ a z -4=0 [ z z =|z|^2 ] |z|^2+a z+( a z )-4=0 [ z ₁ z ₂= z₁ z₂ ] |z|^2+2 Re (a z)-4=0 ( i )[z+ z =2 Re (z)] Now, given z=x+i y |z|^2=x^2+y^2 and a=1+i , then a z=(1+i)(x+i y) a z=(x-y)+i(x+y) Re (a z)=x-y From Eq. (i), put Re (a z)=x-y and |z|^2=x^2+y^2 x^2+y^2+2(x-y)-4=0 S: x^2+y^2+2 x-2 y-4=0 ( ii ) is a circle and given line L: z+ z -i(z- z )+2=0 L: 2 Re (z)-i 2 Im (z)+2=0 L: 2 x-i(2 y i)+2=0 L: x+y+1=0 ( iii ) Now, equation of circle passing through point of intersection of circle S and line L