The line `y=mx+c` cuts the circle `x^2 + y^2 = a^2` at two distinct points A and B. Equation of the circle having minimum radius that can be drawn through the points A and B is:

Options

A. `( 1+ m^(2)) (x^(2) + y^(2) -a^(2)) + 2c ( y -mx- c) = 0`
B. `( 1+ m^(2)) (x^(2) + y^(2) -a^(2)) + c ( y -mx- c) = 0`
C. `( 1+ m^(2)) (x^(2) + y^(2) -a^(2)) -2 c ( y -mx- c) = 0`
D. `( 1+ m^(2)) (x^(2) + y^(2) -a^(2)) - c ( y -mx- c) = 0`

Answer

To solve the problem, we need to find the equation of the circle that has the minimum radius and passes through the points A and B where the line \(y = mx + c\) intersects the circle \(x^2 + y^2 = a^2\).

### Step-by-Step Solution:

1. **Substituting the Line Equation into the Circle Equation**:
We start with the equations:
- Circle: \(x^2 + y^2 = a^2\)
- Line: \(y = mx + c\)

Substitute \(y\) from the line equation into the circle equation:
\[ x^2 + (mx + c)^2 = a^2 \]
Expanding this gives:
\[ x^2 + (m^2x^2 + 2mcx + c^2) = a^2 \]
Rearranging, we have:
\[ (1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0 \]

2. **Finding the Roots**:
Let the roots of the quadratic equation be \(x_1\) and \(x_2\). By Vieta's formulas:
- Sum of roots: \(x_1 + x_2 = -\frac{2mc}{1 + m^2}\)
- Product of roots: \(x_1 x_2 = \frac{c^2 - a^2}{1 + m^2}\)

3. **Finding the Midpoint of A and B**:
The coordinates of the midpoint \(M\) of points A and B are:
\[ M_x = \frac{x_1 + x_2}{2} = -\frac{mc}{1 + m^2} \]
To find \(M_y\), we substitute \(x_1\) and \(x_2\) back into the line equation:
\[ M_y = \frac{y_1 + y_2}{2} = \frac{m(x_1 + x_2) + 2c}{2} = \frac{m\left(-\frac{2mc}{1 + m^2}\right) + 2c}{2} \]
Simplifying this gives:
\[ M_y = \frac{c(1 - m^2)}{1 + m^2} \]

4. **Finding the Radius**:
The distance \(AB\) can be found using the distance formula:
\[ d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
We can express \(y_1 - y_2\) in terms of \(x_1\) and \(x_2\):
\[ y_1 - y_2 = m(x_1 - x_2) \]
Thus,
\[ d = \sqrt{(x_1 - x_2)^2 + m^2(x_1 - x_2)^2} = |x_1 - x_2|\sqrt{1 + m^2} \]
The radius \(r\) of the circle passing through points A and B is:
\[ r = \frac{d}{2} = \frac{|x_1 - x_2|}{2}\sqrt{1 + m^2} \]

5. **Equation of the Circle**:
The equation of the circle with center \(M\) and radius \(r\) is:
\[ \left(x + \frac{mc}{1 + m^2}\right)^2 + \left(y - \frac{c(1 - m^2)}{1 + m^2}\right)^2 = r^2 \]

Substituting \(r^2\) gives:
\[ \left(x + \frac{mc}{1 + m^2}\right)^2 + \left(y - \frac{c(1 - m^2)}{1 + m^2}\right)^2 = \frac{(c^2 - a^2)(1 + m^2)}{4(1 + m^2)} \]

### Final Equation:
The final equation of the circle is:
\[ (1 + m^2)x^2 + (1 + m^2)y^2 - 2cmx - 2c(1 - m^2)y - (c^2 - a^2) = 0 \]