To solve the problem of finding the locus of the center of the circle passing through the origin \( O \) and points \( A \) and \( B \) where the line meets the axes, we can follow these steps:
### Step 1: Understand the Geometry
We have a line at a distance \( c \) from the origin. The line can be represented in the form \( Ax + By = c \). The points where this line intersects the x-axis and y-axis are points \( A \) and \( B \).
### Step 2: Find Points A and B
- The x-intercept \( A \) occurs when \( y = 0 \):
\[
Ax + 0 = c \implies x = \frac{c}{A} \implies A = \left(\frac{c}{A}, 0\right)
\]
- The y-intercept \( B \) occurs when \( x = 0 \):
\[
0 + By = c \implies y = \frac{c}{B} \implies B = \left(0, \frac{c}{B}\right)
\]
### Step 3: Center of the Circle
Let the center of the circle be \( (H, K) \). The circle passes through the origin \( O(0, 0) \), \( A \), and \( B \).
### Step 4: Use the Condition for the Circle
The condition for a point \( (H, K) \) to be the center of a circle passing through points \( O(0, 0) \), \( A\left(\frac{c}{A}, 0\right) \), and \( B\left(0, \frac{c}{B}\right) \) is that the distances from \( (H, K) \) to these points must be equal.
### Step 5: Set Up the Equations
1. Distance from \( (H, K) \) to \( O(0, 0) \):
\[
\sqrt{H^2 + K^2}
\]
2. Distance from \( (H, K) \) to \( A\left(\frac{c}{A}, 0\right) \):
\[
\sqrt{\left(H - \frac{c}{A}\right)^2 + K^2}
\]
3. Distance from \( (H, K) \) to \( B\left(0, \frac{c}{B}\right) \):
\[
\sqrt{H^2 + \left(K - \frac{c}{B}\right)^2}
\]
### Step 6: Equate the Distances
Setting the distances equal gives us two equations:
1. \( H^2 + K^2 = \left(H - \frac{c}{A}\right)^2 + K^2 \)
2. \( H^2 + K^2 = H^2 + \left(K - \frac{c}{B}\right)^2 \)
### Step 7: Simplify the Equations
From the first equation:
\[
H^2 + K^2 = H^2 - 2H\frac{c}{A} + \frac{c^2}{A^2} + K^2
\]
This simplifies to:
\[
0 = -2H\frac{c}{A} + \frac{c^2}{A^2} \implies 2H\frac{c}{A} = \frac{c^2}{A^2} \implies H = \frac{c}{2A}
\]
From the second equation:
\[
H^2 + K^2 = H^2 + K^2 - 2K\frac{c}{B} + \frac{c^2}{B^2}
\]
This simplifies to:
\[
0 = -2K\frac{c}{B} + \frac{c^2}{B^2} \implies 2K\frac{c}{B} = \frac{c^2}{B^2} \implies K = \frac{c}{2B}
\]
### Step 8: Find the Locus
Now substituting \( H \) and \( K \) into the equation:
\[
H^2 + K^2 = \left(\frac{c}{2A}\right)^2 + \left(\frac{c}{2B}\right)^2
\]
This leads to:
\[
H^2 + K^2 = \frac{c^2}{4A^2} + \frac{c^2}{4B^2}
\]
### Step 9: Conclude the Locus
The locus of the center of the circle is given by:
\[
H^2 + K^2 = \frac{c^2}{4} \left(\frac{1}{A^2} + \frac{1}{B^2}\right)
\]
### Final Answer
The locus of the center of the circle passing through \( O, A, B \) is a circle with radius depending on the coefficients of the line.
