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

The equation $x^2 - y^2 - 2x + 4y - 3 = 0$ represents a pair of intersecting lines, which are diameters of the circle. Since $B^2 - 4AC = 0 - 4(1)(-1) = 4 > 0$ and the condition for a pair of straight lines is satisfied, we factor it into two linear equations.

Assume the lines take the form $(x - y + c_1)(x + y + c_2) = 0$. Expanding gives $x^2 - y^2 + (c_1 + c_2)x + (c_1 - c_2)y + c_1c_2 = 0$.

Matching coefficients with the original equation yields:

$c_1 + c_2 = -2$

$c_1 - c_2 = 4$

$c_1c_2 = -3$

Adding the first two equations gives $2c_1 = 2$ so $c_1 = 1$, and substitution shows $c_2 = -3$, consistent with the third equation.

The diameters are thus $x - y + 1 = 0$ and $x + y - 3 = 0$.

Their intersection, the circle's center, is found by solving the system: adding the equations gives $2x - 2 = 0$ so $x = 1$, and substitution into $x + y - 3 = 0$ yields $y = 2$, giving center $(1, 2)$.

The circle passes through $(1, 1)$, so the radius squared is $r^2 = (1 - 1)^2 + (1 - 2)^2 = 1$.

The equation of the circle is $(x - 1)^2 + (y - 2)^2 = 1$, which corresponds to option C.