The equation of the circle passing through the point $(1, 1)$ and having two diameters along the pair of lines ;$x^{2}-y^{2}-2x+4y-3=0$ is

$x^{2}-y^{2}-2x+4y-3=0$

$\Rightarrow x^2-2x+1 = y^2-4y+4$

$\Rightarrow (x-1)^2 = (y-2)^2\Rightarrow x-1 = \pm (y-2)$

Thus lines are $x-y=-1$ and $x+y=3$

Solving these lines we get the centre of the required circle which is $(1, 2)$

Thus radius of the circle is $\sqrt{(1-1)^2+(1-2)^2} = 1$

Hence required circle is $(x-1)^2+(y-2)^2 = 1\Rightarrow x^2+y^2-2x-4y+4=0$