From points on the straight line 3x-4y + 12 = 0, tangents are drawn to the circle `x^2 +y^2 = 4`. Then, the chords of contact pass through a fixed point. The slope of the chord of the circle having this fixed point as its mid-point is

Options

A. `(4)/(3)`
B. `(1)/(2)`
C. `(1)/(3)`
D. none of these

Explanation

To solve the problem, we need to find the slope of the chord of the circle \(x^2 + y^2 = 4\) that has a fixed point as its midpoint. The fixed point is derived from the tangents drawn from points on the line \(3x - 4y + 12 = 0\).

### Step-by-Step Solution:

1. **Identify the line equation**: The equation of the line is given as \(3x - 4y + 12 = 0\). We can rearrange it to find \(y\) in terms of \(x\):
\[ 4y = 3x + 12 \implies y = \frac{3}{4}x + 3 \]

2. **Choose a point on the line**: Let’s take a point \(P(\alpha, \frac{3\alpha + 12}{4})\) on the line, where \(\alpha\) is any real number.

3. **Equation of the chord of contact**: The equation of the chord of contact from point \(P(x_1, y_1)\) to the circle \(x^2 + y^2 = 4\) is given by:
\[ xx_1 + yy_1 = 4 \]
Substituting \(x_1 = \alpha\) and \(y_1 = \frac{3\alpha + 12}{4}\):
\[ x\alpha + y\left(\frac{3\alpha + 12}{4}\right) = 4 \]

4. **Simplify the equation**: Multiply through by 4 to eliminate the fraction:
\[ 4x\alpha + y(3\alpha + 12) = 16 \]
Rearranging gives:
\[ 4\alpha x + (3\alpha + 12)y - 16 = 0 \]

5. **Identify the fixed point**: The problem states that the chords of contact pass through a fixed point. We need to find this fixed point. The fixed point \(P\) is given as \((-1, \frac{4}{3})\).

6. **Substituting the fixed point into the chord equation**: To find the slope of the chord, we can substitute the coordinates of the fixed point into the equation of the chord:
\[ 4\alpha(-1) + (3\alpha + 12)\left(\frac{4}{3}\right) - 16 = 0 \]
Simplifying this:
\[ -4\alpha + \frac{4(3\alpha + 12)}{3} - 16 = 0 \]
\[ -4\alpha + 4\alpha + 16 - 16 = 0 \]
This shows that the equation holds true for any \(\alpha\), confirming that the chords pass through the fixed point.

7. **Finding the slope of the chord**: The slope of the line joining the center of the circle (0, 0) to the fixed point \((-1, \frac{4}{3})\) is given by:
\[ \text{slope} = \frac{\frac{4}{3} - 0}{-1 - 0} = -\frac{4}{3} \]

8. **Conclusion**: The slope of the chord of the circle having the fixed point as its midpoint is \(-\frac{4}{3}\).