Step 1: Understanding the Concept:
The diameters of a circle always intersect at its center.
The given equation represents a pair of straight lines. Since these lines are diameters, their point of intersection will be the center of the circle.
Once the center is found, the radius can be calculated as the distance from the center to the given point $(1, 1)$.
Step 2: Key Formula or Approach:
Factorize the combined equation of lines to find the individual lines and their intersection.
Alternatively, to find the intersection point $(h, k)$ of a pair of lines $ax^2+2hxy+by^2+2gx+2fy+c=0$, we can take partial derivatives: $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
Radius $r = \sqrt{(x - h)^2 + (y - k)^2}$.
Equation of circle is $(x - h)^2 + (y - k)^2 = r^2$.
Step 3: Detailed Explanation:
Let's find the center of the circle by finding the intersection of the pair of lines:
$f(x, y) = x^2 - y^2 - 2x + 4y - 3 = 0$
Using partial derivatives to find the intersection point:
$\frac{\partial f}{\partial x} = 2x - 2 = 0 \Rightarrow x = 1$
$\frac{\partial f}{\partial y} = -2y + 4 = 0 \Rightarrow y = 2$
So, the center of the circle is $(h, k) = (1, 2)$.
We can also verify this by completing squares to factorize the equation:
$(x^2 - 2x) - (y^2 - 4y) - 3 = 0$
$(x^2 - 2x + 1) - (y^2 - 4y + 4) - 3 - 1 + 4 = 0$
$(x - 1)^2 - (y - 2)^2 = 0$
$((x - 1) - (y - 2)) ((x - 1) + (y - 2)) = 0$
$(x - y + 1)(x + y - 3) = 0$
The lines are $x - y + 1 = 0$ and $x + y - 3 = 0$. Solving these gives $x = 1, y = 2$.
The circle passes through the point $(1, 1)$.
Radius $r = \text{distance from center to point}$
$r = \sqrt{(1 - 1)^2 + (1 - 2)^2} = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1$.
The equation of the circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
\[ (x - 1)^2 + (y - 2)^2 = 1^2 \]
\[ (x - 1)^2 + (y - 2)^2 = 1 \]
Step 4: Final Answer:
The equation of the circle is $(x - 1)^2 + (y - 2)^2 = 1$.