Key Concepts and Formulas

• Equation of a circle touching the line $Ax + By + C = 0$ at point $(x_1, y_1)$: $(x - x_1)^2 + (y - y_1)^2 + \lambda (Ax + By + C) = 0$
• General equation of a circle: $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$

Step-by-Step Solution

1. Formulate the General Equation of the Circle

• State what you are doing and why: We want to find the general equation of the circle that touches the line $x = y$ at the point $(1, 1)$. We use the formula for a circle touching a line at a given point.
• Show the math: The given point is $(x_1, y_1) = (1, 1)$, and the tangent line is $x = y$, which can be rewritten as $x - y = 0$. Substituting into the formula, we get: $(x - 1)^2 + (y - 1)^2 + \lambda (x - y) = 0 \quad (*)$
• Explain the reasoning: This equation represents a family of circles that are tangent to the line $x=y$ at the point $(1,1)$. The parameter $\lambda$ will be determined by the additional condition that the circle passes through $(1,-3)$.

2. Determine the Value of $\lambda$

• State what you are doing and why: We use the condition that the circle passes through the point $(1, -3)$ to find the specific value of $\lambda$.
• Show the math: Substituting $x = 1$ and $y = -3$ into equation $(*)$, we get: $(1 - 1)^2 + (-3 - 1)^2 + \lambda (1 - (-3)) = 0$ $0 + (-4)^2 + \lambda (4) = 0$ $16 + 4\lambda = 0$
• Explain the reasoning: Substituting the coordinates of the point into the equation of the circle gives us an equation that can be solved for $\lambda$.
• Show the math: Solving for $\lambda$: $4\lambda = -16$ $\lambda = -4$

3. Write the Specific Equation of the Circle

• State what you are doing and why: We substitute the value of $\lambda$ back into the equation $(*)$ to get the equation of the specific circle.
• Show the math: Substituting $\lambda = -4$ into equation $(*)$: $(x - 1)^2 + (y - 1)^2 - 4 (x - y) = 0$
• Explain the reasoning: This is the equation of the circle that satisfies both the tangency condition and the condition of passing through the point $(1,-3)$.
• Show the math: Expanding and simplifying: $(x^2 - 2x + 1) + (y^2 - 2y + 1) - 4x + 4y = 0$ $x^2 + y^2 - 6x + 2y + 2 = 0$

4. Calculate the Radius of the Circle

• State what you are doing and why: We use the general form of the circle equation to find the radius.
• Show the math: Comparing $x^2 + y^2 - 6x + 2y + 2 = 0$ with $x^2 + y^2 + 2gx + 2fy + c = 0$:
  • $2g = -6 \Rightarrow g = -3$
  • $2f = 2 \Rightarrow f = 1$
  • $c = 2$
• Explain the reasoning: Comparing the equation with the general form allows us to identify the values of $g$, $f$, and $c$, which are used to calculate the radius.
• Show the math: The radius is given by: $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (1)^2 - 2} = \sqrt{9 + 1 - 2} = \sqrt{8} = 2\sqrt{2}$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when substituting and simplifying equations.
• Incorrect Lambda: A positive lambda was initially obtained. Always double-check your substitution and arithmetic.
• Radius Formula: Remember the radius formula is $\sqrt{g^2 + f^2 - c}$, not just $\sqrt{g^2 + f^2}$.

Summary

We first set up the general equation of the circle using the condition that it touches the line $x = y$ at the point $(1, 1)$. Then, we used the fact that the circle also passes through the point $(1, -3)$ to determine the specific value of the parameter $\lambda$. Substituting this value back into the equation, we obtained the equation of the circle in the general form, from which we calculated the radius to be $2\sqrt{2}$.

The final answer is $\boxed{2\sqrt{2}}$, which corresponds to option (C).