Key Concepts and Formulas

• Equation of a Circle: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$. The standard form is $(x-h)^2 + (y-k)^2 = r^2$, with center $(h,k)$ and radius $r$.
• Mirror Image of a Point in a Line: The mirror image of a point $(x_1, y_1)$ in the line $ax + by + c = 0$ is $(x_2, y_2)$, where $\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$.
• The midpoint of the point and its image lies on the line.

Step-by-Step Solution

Step 1: Analyze the equation of circle $C_1$ and find its center and radius.

The equation of circle $C_1$ is given by $x^2 + y^2 - 4x - 2y = \alpha - 5$. We rewrite this in the standard form by completing the square: $(x^2 - 4x) + (y^2 - 2y) = \alpha - 5$ $(x^2 - 4x + 4) + (y^2 - 2y + 1) = \alpha - 5 + 4 + 1$ $(x - 2)^2 + (y - 1)^2 = \alpha$ Thus, the center of $C_1$ is $(2, 1)$ and its radius is $\sqrt{\alpha}$. We require $\alpha > 0$ for the circle to be real.

Step 2: Analyze the equation of circle $C_2$ and find its center and radius.

The equation of circle $C_2$ is given by $5x^2 + 5y^2 - 10fx - 10gy + 36 = 0$. Divide by 5 to get $x^2 + y^2 - 2fx - 2gy + \frac{36}{5} = 0$ The center of $C_2$ is $(f, g)$ and its radius is $r = \sqrt{f^2 + g^2 - \frac{36}{5}}$.

Step 3: Find the mirror image of the center of $C_1$ in the line $y = 2x + 1$.

The line is $2x - y + 1 = 0$. Let the mirror image of the center $(2, 1)$ be $(f, g)$. Using the mirror image formula: $\frac{f - 2}{2} = \frac{g - 1}{-1} = \frac{-2(2(2) - 1 + 1)}{2^2 + (-1)^2} = \frac{-2(4)}{5} = -\frac{8}{5}$ Then, $f = 2 + 2\left(-\frac{8}{5}\right) = 2 - \frac{16}{5} = \frac{10 - 16}{5} = -\frac{6}{5}$ $g = 1 - 1\left(-\frac{8}{5}\right) = 1 + \frac{8}{5} = \frac{5 + 8}{5} = \frac{13}{5}$ So, the center of $C_2$ is $\left(-\frac{6}{5}, \frac{13}{5}\right)$.

Step 4: Relate the radii of $C_1$ and $C_2$.

Since $C_2$ is the mirror image of $C_1$, their radii are equal. Therefore, $r = \sqrt{\alpha}$. Also, we know that the radius of $C_2$ is $r = \sqrt{f^2 + g^2 - \frac{36}{5}}$. Substitute the values of $f$ and $g$: $r = \sqrt{\left(-\frac{6}{5}\right)^2 + \left(\frac{13}{5}\right)^2 - \frac{36}{5}} = \sqrt{\frac{36}{25} + \frac{169}{25} - \frac{180}{25}} = \sqrt{\frac{36 + 169 - 180}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1$ Thus, $r = 1$.

Step 5: Find the value of $\alpha$.

Since $r = \sqrt{\alpha}$ and $r = 1$, we have $\sqrt{\alpha} = 1$, which implies $\alpha = 1$.

Step 6: Calculate $\alpha + r$.

We have $\alpha = 1$ and $r = 1$. Therefore, $\alpha + r = 1 + 1 = 2$.

Step 7: Sanity Check The midpoint of $(2, 1)$ and $(-\frac{6}{5}, \frac{13}{5})$ must lie on the line $y = 2x + 1$. The midpoint is $(\frac{2 - 6/5}{2}, \frac{1 + 13/5}{2}) = (\frac{4/5}{2}, \frac{18/5}{2}) = (\frac{2}{5}, \frac{9}{5})$. Check if $(\frac{2}{5}, \frac{9}{5})$ lies on $y = 2x + 1$: $\frac{9}{5} = 2(\frac{2}{5}) + 1 = \frac{4}{5} + 1 = \frac{9}{5}$. This is correct!

Step 8: Recalculate The correct answer is given as 1. Let's re-examine the problem. The center of $C_1$ is $(2, 1)$ and its radius is $\sqrt{\alpha}$. The center of $C_2$ is $(f, g) = (-\frac{6}{5}, \frac{13}{5})$ and its radius is $r = \sqrt{f^2 + g^2 - \frac{36}{5}} = 1$. The problem states that $5x^2 + 5y^2 - 10fx - 10gy + 36 = 0$ represents the reflection of $C_1$ in the line $y = 2x + 1$. Since the radius of $C_2$ is 1, the radius of $C_1$ is also 1. Therefore, $\sqrt{\alpha} = 1$, so $\alpha = 1$. We are asked to find $\alpha + r = 1 + 1 = 2$.

The given answer is 1, so we must be wrong. Let's check the calculations again. The radius of $C_2$ is 1. Thus, $r = 1$. So, we have $r^2 = 1$. Also, $r^2 = f^2 + g^2 - \frac{36}{5}$. We also have $\alpha = r^2 = 1$. Therefore, $\alpha + r = 1 + r$.

The equation of $C_1$ is $x^2 + y^2 - 4x - 2y = \alpha - 5$. Comparing this with $x^2 + y^2 + 2Gx + 2Fy + C = 0$, we have $2G = -4$ and $2F = -2$. Thus, $G = -2$ and $F = -1$. The center of $C_1$ is $(2, 1)$ and radius is $\sqrt{G^2 + F^2 - C} = \sqrt{(-2)^2 + (-1)^2 - (\alpha - 5)} = \sqrt{4 + 1 - \alpha + 5} = \sqrt{10 - \alpha}$. Since $r = \sqrt{10 - \alpha} = \sqrt{\alpha}$, we have $10 - \alpha = \alpha$, which gives $2\alpha = 10$, so $\alpha = 5$. $r = \sqrt{5}$. Thus, $\alpha + r = 5 + \sqrt{5}$. Still not getting 1.

Let's reconsider the radius of $C_1$. We have $(x-2)^2 + (y-1)^2 = \alpha$. Hence, $r_1 = \sqrt{\alpha}$. Since $r_1 = r = 1$, we have $\sqrt{\alpha} = 1$, so $\alpha = 1$. Therefore, $\alpha + r = 1 + 1 = 2$.

Final Attempt We have $C_1: (x-2)^2 + (y-1)^2 = \alpha$, center $(2, 1)$, radius $\sqrt{\alpha}$. $C_2: x^2 + y^2 - 2fx - 2gy + \frac{36}{5} = 0$, center $(f, g)$, radius $\sqrt{f^2 + g^2 - \frac{36}{5}} = r$. Since $C_2$ is the mirror image of $C_1$, their radii are equal, so $\sqrt{\alpha} = r$. Thus, $\alpha = r^2$. The center of $C_2$ is the mirror image of $(2, 1)$ in the line $2x - y + 1 = 0$. So the center of $C_2$ is $(-\frac{6}{5}, \frac{13}{5})$. Thus, $f = -\frac{6}{5}$ and $g = \frac{13}{5}$. Then $r = \sqrt{\frac{36}{25} + \frac{169}{25} - \frac{36}{5}} = \sqrt{\frac{36 + 169 - 180}{25}} = \sqrt{\frac{25}{25}} = 1$. Therefore, $r = 1$. Then $\alpha = r^2 = 1^2 = 1$. Finally, $\alpha + r = 1 + 1 = 2$.

I have been unable to arrive at the correct answer of 1. I suspect there may be an error in the provided solution or the given correct answer.

Common Mistakes & Tips

• Be careful while finding the image of a point in a line. Ensure the formula is applied correctly.
• Remember that the radius of a circle must be a positive real number. This places restrictions on parameters like $\alpha$.
• Double-check your calculations, especially when dealing with fractions and square roots.

Summary

We found the centers and radii of both circles. We used the mirror image formula to relate the center of $C_1$ to the center of $C_2$. We equated the radii of the two circles and solved for $\alpha$. Finally, we calculated $\alpha + r$. However, the calculations lead to $\alpha + r = 2$, which contradicts the given answer of 1.

The final answer is \boxed{2}. There is no correct option.