Key Concepts and Formulas

• Equation of a Circle with Diameter Endpoints: If $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a diameter, the equation of the circle is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
• Image of a Point in a Line: The image $(x', y')$ of a point $(x_1, y_1)$ in the line $Ax+By+C=0$ is given by the formula: $\frac{x' - x_1}{A} = \frac{y' - y_1}{B} = \frac{-2(Ax_1+By_1+C)}{A^2+B^2}$
• General Equation of a Circle: A circle with center $(h,k)$ and radius $r$ has the equation $(x-h)^2 + (y-k)^2 = r^2$, which expands to $x^2+y^2-2hx-2ky+h^2+k^2-r^2=0$.

Step-by-Step Solution

Step 1: Finding the Points of Intersection A and B

We are given the line $ax + by = 0$ (where $a \ne b$) and the circle $x^2 + y^2 - 2x = 0$. The points of intersection are $A(\alpha, 0)$ and $B(1, \beta)$. Our goal is to find the values of $\alpha$ and $\beta$.

• Finding $\alpha$: Since $A(\alpha, 0)$ lies on the line $ax + by = 0$, we have $a\alpha + b(0) = 0 \implies a\alpha = 0$. Since the problem does not specify that $a=0$, and to avoid the degenerate case of $y=0$, we assume $a \ne 0$, implying $\alpha = 0$. Thus, $A(0, 0)$.

• Finding $\beta$: Since $B(1, \beta)$ lies on the line $ax + by = 0$, we have $a(1) + b\beta = 0 \implies a + b\beta = 0 \implies \beta = -\frac{a}{b}$. Since $B(1, \beta)$ also lies on the circle $x^2 + y^2 - 2x = 0$, we have $1^2 + \beta^2 - 2(1) = 0 \implies 1 + \beta^2 - 2 = 0 \implies \beta^2 = 1 \implies \beta = \pm 1$.

  If $\beta = 1$, then $1 = -\frac{a}{b} \implies a = -b$. If $\beta = -1$, then $-1 = -\frac{a}{b} \implies a = b$. Since we are given $a \ne b$, we must have $a = -b$ and $\beta = 1$. Therefore, $B(1, 1)$.

Step 2: Determining the Original Circle with AB as Diameter

We have the diameter endpoints $A(0, 0)$ and $B(1, 1)$. We want to find the equation of the circle with AB as its diameter.

The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$. Substituting $A(0,0)$ and $B(1,1)$, we get: $(x-0)(x-1) + (y-0)(y-1) = 0$ $x(x-1) + y(y-1) = 0$ $x^2 - x + y^2 - y = 0$ $x^2 + y^2 - x - y = 0$

From the equation $x^2 + y^2 - x - y = 0$, we can find the center and radius. Comparing with the general form $x^2+y^2+2gx+2fy+c=0$, we have $2g = -1 \implies g = -\frac{1}{2}$ and $2f = -1 \implies f = -\frac{1}{2}$. The center is $(-g, -f) = (\frac{1}{2}, \frac{1}{2})$. The radius $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{1}{2})^2 + (-\frac{1}{2})^2 - 0} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Step 3: Finding the Image of the Circle's Center

The circle with AB as diameter has center $C_1(\frac{1}{2}, \frac{1}{2})$ and radius $r = \frac{1}{\sqrt{2}}$. We need to find the image of this center in the line $x + y + 2 = 0$.

Let the image of $C_1(\frac{1}{2}, \frac{1}{2})$ in the line $x + y + 2 = 0$ be $C_2(h, k)$. Using the point reflection formula: $\frac{h - \frac{1}{2}}{1} = \frac{k - \frac{1}{2}}{1} = \frac{-2(1(\frac{1}{2}) + 1(\frac{1}{2}) + 2)}{1^2 + 1^2} = \frac{-2(\frac{1}{2} + \frac{1}{2} + 2)}{2} = \frac{-2(3)}{2} = -3$

So, $h - \frac{1}{2} = -3 \implies h = -3 + \frac{1}{2} = -\frac{5}{2}$ and $k - \frac{1}{2} = -3 \implies k = -3 + \frac{1}{2} = -\frac{5}{2}$. The image of the center is $C_2(-\frac{5}{2}, -\frac{5}{2})$.

Step 4: Finding the Equation of the Image Circle

The image circle has center $C_2(-\frac{5}{2}, -\frac{5}{2})$ and the same radius $r = \frac{1}{\sqrt{2}}$. The equation of the image circle is: $(x - h)^2 + (y - k)^2 = r^2$ $(x + \frac{5}{2})^2 + (y + \frac{5}{2})^2 = (\frac{1}{\sqrt{2}})^2$ $x^2 + 5x + \frac{25}{4} + y^2 + 5y + \frac{25}{4} = \frac{1}{2}$ $x^2 + y^2 + 5x + 5y + \frac{25}{4} + \frac{25}{4} - \frac{1}{2} = 0$ $x^2 + y^2 + 5x + 5y + \frac{50}{4} - \frac{2}{4} = 0$ $x^2 + y^2 + 5x + 5y + \frac{48}{4} = 0$ $x^2 + y^2 + 5x + 5y + 12 = 0$

Common Mistakes & Tips

• Assuming a=0: Remember that while $a\alpha = 0$ implies either $a=0$ or $\alpha=0$, the problem context often restricts $a$ from being zero to avoid a trivial line equation.
• Sign errors in reflection formula: Be careful with the signs in the image formula. A common mistake is to forget the $-2$ factor.
• Forgetting to square the radius: When constructing the equation of the circle, remember to square the radius in the equation $(x-h)^2 + (y-k)^2 = r^2$.

Summary

We first found the intersection points A and B of the given line and circle. Using A and B as diameter endpoints, we determined the equation of the original circle. We then found the image of the center of this circle in the given line, keeping the radius constant. Finally, we wrote the equation of the image circle, which is $x^2 + y^2 + 5x + 5y + 12 = 0$. This corresponds to option (A).

Final Answer

The final answer is \boxed{{x^2} + {y^2} + 5x + 5y + 12 = 0}, which corresponds to option (A).