JEE Circles 2

Any circle through points (t,1) and (1,t) can be given by

(x-1)(x-t) + (y-1)(y-t) +k(line equation through (1,t) and (t,1))=0

Since this circle also passes through (t,t) We find that the circle equation is

(x-1)(x-t) +(y-1)(y-t)=0

From this equation it can easily be seen that the fixed point is (1,1)

Isn't the argument of any fixed point pi/4?

Let $t=0$ . Then, the circle passes through $(1,0)$ , $(0,1)$ and $(0,0)$ . By symmetry, the circle will also pass through $(1,1)$ . Then,

$arg(1+i)=\dfrac{\pi}{4}$

draw the circle and you see that the circle passes through $(1,1)$ ,
so $a=1,b=1$ , then argument will be $tan^{-1}\dfrac{a}{b} = tan^{-1}1=\dfrac{\pi}{4}$