The points of intersection of the line ax + by = 0, `(a != b) ` and the circle `x^2 + y^2 – 2x = 0` are` A (alpha , 0)` and `B (1, beta ).` The image of the circle with AB as a diameter in the line x + y + 2 = 0 is :

Options

A. `x^2+ y^2+ 3x + 3y + 4 = 0`
B. `x^2 + y^2 + 5x + 5y + 12 = 0`
C. `x^2 + y^2 + 3x + 5y + 8 = 0`
D. `x^2+ y^2 – 5x – 5y + 12 = 0`

Answer

To solve the problem, we need to find the image of the circle with diameter AB, where A and B are the points of intersection of the line \( ax + by = 0 \) and the circle \( x^2 + y^2 - 2x = 0 \), in the line \( x + y + 2 = 0 \).

### Step 1: Find the points of intersection A and B

The line is given by \( ax + by = 0 \). We can express \( y \) in terms of \( x \):
\[ y = -\frac{a}{b}x \]

The circle can be rewritten as:
\[ x^2 + y^2 - 2x = 0 \implies x^2 + y^2 = 2x \]

Substituting \( y = -\frac{a}{b}x \) into the circle's equation:
\[ x^2 + \left(-\frac{a}{b}x\right)^2 = 2x \]
\[ x^2 + \frac{a^2}{b^2}x^2 = 2x \]
\[ \left(1 + \frac{a^2}{b^2}\right)x^2 - 2x = 0 \]
Factoring out \( x \):
\[ x\left( \left(1 + \frac{a^2}{b^2}\right)x - 2 \right) = 0 \]
This gives us \( x = 0 \) or \( x = \frac{2b^2}{b^2 + a^2} \).

For \( x = 0 \):
\[ y = -\frac{a}{b}(0) = 0 \implies A(0, 0) \]

For \( x = \frac{2b^2}{b^2 + a^2} \):
\[ y = -\frac{a}{b}\left(\frac{2b^2}{b^2 + a^2}\right) = -\frac{2ab^2}{b^2 + a^2} \implies B\left(\frac{2b^2}{b^2 + a^2}, -\frac{2ab^2}{b^2 + a^2}\right) \]

### Step 2: Find the center and radius of the circle with diameter AB

The center \( C \) of the circle with diameter AB is the midpoint of A and B:
\[ C\left(\frac{0 + \frac{2b^2}{b^2 + a^2}}{2}, \frac{0 - \frac{2ab^2}{b^2 + a^2}}{2}\right) = \left(\frac{b^2}{b^2 + a^2}, -\frac{ab^2}{b^2 + a^2}\right) \]

The radius \( R \) is half the distance between A and B. The distance \( d \) between A and B is:
\[ d = \sqrt{\left(\frac{2b^2}{b^2 + a^2} - 0\right)^2 + \left(-\frac{2ab^2}{b^2 + a^2} - 0\right)^2} \]
Calculating \( d \):
\[ d = \sqrt{\left(\frac{2b^2}{b^2 + a^2}\right)^2 + \left(-\frac{2ab^2}{b^2 + a^2}\right)^2} \]
\[ = \sqrt{\frac{4b^4 + 4a^2b^4}{(b^2 + a^2)^2}} = \frac{2b^2\sqrt{1 + a^2}}{b^2 + a^2} \]
Thus, the radius \( R \) is:
\[ R = \frac{d}{2} = \frac{b^2\sqrt{1 + a^2}}{b^2 + a^2} \]

### Step 3: Find the image of the center C about the line \( x + y + 2 = 0 \)

To find the image of point \( C \) about the line \( x + y + 2 = 0 \), we use the formula for the reflection of a point \( (x_1, y_1) \) about the line \( Ax + By + C = 0 \):
\[ \text{Image} = \left( x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2}, y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \right) \]
Here, \( A = 1, B = 1, C = 2 \), and \( C\left(\frac{b^2}{b^2 + a^2}, -\frac{ab^2}{b^2 + a^2}\right) \).

Calculating \( Ax_1 + By_1 + C \):
\[ = 1\cdot\frac{b^2}{b^2 + a^2} + 1\cdot\left(-\frac{ab^2}{b^2 + a^2}\right) + 2 = \frac{b^2 - ab^2 + 2(b^2 + a^2)}{b^2 + a^2} = \frac{(1 - a + 2)b^2 + 2a^2}{b^2 + a^2} \]

Now substituting into the reflection formula gives us the coordinates of the image.

### Final Step: Write the equation of the image circle

The image circle will have the same radius \( R \) and the center as calculated above. The equation of the circle is:
\[ \left(x - x_C\right)^2 + \left(y - y_C\right)^2 = R^2 \]