A circle passes through the origin O and cuts the axes at A (a,0) and B (0,b). The image of origin O in the line AB is the point

Options

A. `((2ab ^(2))/( a ^(2) + b ^(2)), (2a ^(2) b )/( a ^(2) + b ^(2)))`
B. `((2a ^(2) b )/( a ^(2) + b ^(2)), (2ab ^(2))/(a ^(2) + b ^(2)))`
C. `((2ab )/(a ^(2) +b ^(2)), (2ab )/( a ^(2) + b ^(2)))`
D. `(a,b)`

Answer

To find the image of the origin \( O(0,0) \) in the line \( AB \) where the circle intersects the axes at points \( A(a, 0) \) and \( B(0, b) \), we will follow these steps:

### Step 1: Write the equation of line \( AB \)
The line \( AB \) can be expressed in intercept form as:

\[ \frac{x}{a} + \frac{y}{b} = 1 \]

This can be rearranged to standard form:

\[ bx + ay - ab = 0 \]

### Step 2: Identify coefficients
From the equation \( bx + ay - ab = 0 \), we identify the coefficients:

- \( A = b \)
- \( B = a \)
- \( C = -ab \)

### Step 3: Use the reflection formula
To find the image \( O' \) of the origin \( O(0,0) \) in the line \( AB \), we use the reflection formula:

\[ x' = x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2} \]
\[ y' = y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \]

Substituting \( (x_1, y_1) = (0, 0) \):

\[ x' = 0 - \frac{2b(0 + 0 - ab)}{b^2 + a^2} = \frac{2ab}{a^2 + b^2} \]
\[ y' = 0 - \frac{2a(0 + 0 - ab)}{b^2 + a^2} = \frac{2a^2b}{a^2 + b^2} \]

### Step 4: Write the coordinates of the image point
Thus, the coordinates of the image point \( O' \) are:

\[ O' \left( \frac{2ab}{a^2 + b^2}, \frac{2a^2b}{a^2 + b^2} \right) \]

### Final Answer
The image of the origin \( O \) in the line \( AB \) is:

\[ O' \left( \frac{2ab}{a^2 + b^2}, \frac{2a^2b}{a^2 + b^2} \right) \]

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