The locus of the centre of the circle passing through the origin O and the points of intersection A and B of any line through (a, b) and the coordinate axes is

Options

A. `(x)/(a)+(y)/(b)=1`
B. `(a)/(x)+(b)/(y)=1`
C. `(x)/(a)+(y)/(b)=2`
D. `(a)/(x)+(b)/(y)=2`

Explanation

To find the locus of the center of a circle that passes through the origin \( O(0, 0) \) and the points of intersection \( A \) and \( B \) of any line through \( (a, b) \) and the coordinate axes, we can follow these steps:

### Step 1: Identify the points of intersection
Let the line passing through \( (a, b) \) intersect the x-axis at point \( A(r, 0) \) and the y-axis at point \( B(0, s) \). The coordinates of these points are determined by the line equation.

### Step 2: Write the equation of the line
The equation of the line can be expressed in intercept form as:
\[ \frac{x}{r} + \frac{y}{s} = 1 \]
This line passes through the point \( (a, b) \), so substituting \( (a, b) \) into the line equation gives:
\[ \frac{a}{r} + \frac{b}{s} = 1 \]

### Step 3: Express \( r \) and \( s \) in terms of \( h \) and \( k \)
The center of the circle \( C(h, k) \) is the midpoint of the diameter \( AB \). Thus, we can express \( r \) and \( s \) in terms of \( h \) and \( k \):
\[ r = 2h \quad \text{and} \quad s = 2k \]

### Step 4: Substitute \( r \) and \( s \) into the line equation
Substituting \( r \) and \( s \) into the line equation gives:
\[ \frac{a}{2h} + \frac{b}{2k} = 1 \]

### Step 5: Simplify the equation
Multiplying through by \( 2hk \) to eliminate the denominators results in:
\[ ak + bh = 2hk \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ ak + bh - 2hk = 0 \]

### Step 7: Identify the locus
This equation represents the locus of the center \( C(h, k) \) of the circle.

### Final Equation
Thus, the locus of the center of the circle passing through the origin and points \( A \) and \( B \) is given by:
\[ ak + bh = 2hk \]