Key Concepts and Formulas

• General Equation of a Circle: The standard form of a circle's equation is $x^2 + y^2 + 2gx + 2fy + c = 0$. Its center is at $(-g, -f)$ and its radius is $R = \sqrt{g^2 + f^2 - c}$.
• Reflection of a Point Across a Line: The mirror image $(x', y')$ of a point $(x_1, y_1)$ across the line $ax + by + c = 0$ is given by: $\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$
• Reflection of a Circle: When a circle is reflected across a line, its radius remains unchanged, while its center is reflected to a new location.

Step-by-Step Solution

1. Analyze Circle $c_1$

The equation of circle $c_1$ is $x^2 + y^2 - 2x - 6y + \alpha = 0$. We need to find its center and radius. Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$: $2g = -2 \Rightarrow g = -1$ $2f = -6 \Rightarrow f = -3$ $c = \alpha$

Therefore, the center of $c_1$, denoted as $C_1$, is $(-g, -f) = (1, 3)$. The radius of $c_1$, denoted as $R_1$, is $\sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-3)^2 - \alpha} = \sqrt{1 + 9 - \alpha} = \sqrt{10 - \alpha}$.

Explanation: We extract the center and radius of $c_1$ because for a reflection, the center of the original circle is reflected to become the center of the image circle, while the radius remains the same.*

2. Find the Image of the Center $C_1$ in the Line $y = x + 1$

The line of reflection is $y = x + 1$, which can be rewritten as $x - y + 1 = 0$. Here, $a = 1$, $b = -1$, and $c = 1$. The point to be reflected is $C_1 = (x_1, y_1) = (1, 3)$.

Let the image of $C_1$ be $C_2 = (x', y')$. Using the reflection formula: $\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$ Substituting the values: $\frac{x' - 1}{1} = \frac{y' - 3}{-1} = -2 \frac{(1)(1) + (-1)(3) + 1}{1^2 + (-1)^2}$ $\frac{x' - 1}{1} = \frac{y' - 3}{-1} = -2 \frac{1 - 3 + 1}{1 + 1}$ $\frac{x' - 1}{1} = \frac{y' - 3}{-1} = -2 \frac{-1}{2}$ $\frac{x' - 1}{1} = \frac{y' - 3}{-1} = 1$

Now, we solve for $x'$ and $y'$: $x' - 1 = 1 \Rightarrow x' = 2$ $y' - 3 = -1 \Rightarrow y' = 2$

So, the center of the image circle $c_2$, denoted as $C_2$, is $(2, 2)$.

Explanation: The image of a circle is found by reflecting its center across the line. The coordinates $(2,2)$ will be the center of circle $c_2$.*

3. Analyze Circle $c_2$ and Relate its Properties

The equation of circle $c_2$ is $5x^2 + 5y^2 + 10gx + 10fy + 38 = 0$.

• Normalize the equation: Divide the entire equation by 5: $x^2 + y^2 + 2gx + 2fy + \frac{38}{5} = 0$

• Identify its center: The center of $c_2$ is $(-g, -f)$. We know from Step 2 that the center of $c_2$ is $C_2 = (2, 2)$. Therefore, $(-g, -f) = (2, 2)$, which implies $g = -2$ and $f = -2$.

• Identify its radius: The radius of $c_2$, denoted as $r$, is given by $\sqrt{g^2 + f^2 - \frac{38}{5}}$. Substituting $g = -2$ and $f = -2$: $r = \sqrt{(-2)^2 + (-2)^2 - \frac{38}{5}} = \sqrt{4 + 4 - \frac{38}{5}} = \sqrt{8 - \frac{38}{5}}$ $r = \sqrt{\frac{40 - 38}{5}} = \sqrt{\frac{2}{5}}$

• Equate radii: The radius of the reflected circle is equal to the radius of the original circle. So, $R_1 = r$. $\sqrt{10 - \alpha} = \sqrt{\frac{2}{5}}$ Squaring both sides: $10 - \alpha = \frac{2}{5}$

Explanation: By normalizing the equation of $c_2$, we can easily find its center and radius in terms of $g, f$. Since $c_2$ is the image of $c_1$, its center must be the image of $C_1$, which we found to be $(2,2)$. This allows us to determine the values of $g$ and $f$ for $c_2$. Crucially, the radius of $c_2$ must be the same as the radius of $c_1$. This equality provides an equation to solve for $\alpha$.*

4. Solve for $\alpha$ and $r$

From the radius equality: $10 - \alpha = \frac{2}{5}$ $\alpha = 10 - \frac{2}{5}$ $\alpha = \frac{50 - 2}{5} = \frac{48}{5}$

The radius of circle $c_2$ is $r = \sqrt{\frac{2}{5}}$. Thus, $r^2 = \frac{2}{5}$.

5. Calculate the Final Expression $\alpha + 6r^2$

Now we substitute the values of $\alpha$ and $r^2$: $\alpha + 6r^2 = \frac{48}{5} + 6 \left(\frac{2}{5}\right)$ $= \frac{48}{5} + \frac{12}{5}$ $= \frac{48 + 12}{5}$ $= \frac{60}{5}$ $= 12$

Common Mistakes & Tips

• Normalization: Always normalize the circle equation ($x^2$ and $y^2$ coefficients must be 1) before extracting the center and radius.
• Reflection Formula: Be careful with signs and the correct substitution of $a, b, c, x_1,$ and $y_1$ in the reflection formula.

Summary

This problem involves finding the mirror image of a circle across a line. We first find the center and radius of the original circle. Then, we find the image of the center across the given line. Using the fact that the radius remains unchanged during reflection, we can relate the parameters of the original and image circles and solve for the unknown $\alpha$. Finally, we calculate the required expression.

The final answer is \boxed{12}.