If a circle passes through the points where the lines `3kx- 2y-1 = 0` and `4x-3y + 2 = 0` meet the coordinate axes then `k=`

Options

A. 1
B. `-1`
C. `(1)/(2)`
D. `(-1)/(2)`

Explanation

To solve the problem, we need to find the value of \( k \) such that the circle passes through the points where the lines \( 3kx - 2y - 1 = 0 \) and \( 4x - 3y + 2 = 0 \) intersect the coordinate axes.

### Step 1: Find the intersection points of the lines with the coordinate axes.

**For the line \( 3kx - 2y - 1 = 0 \):**

- **To find the x-intercept (set \( y = 0 \)):**
\[ 3kx - 1 = 0 \implies x = \frac{1}{3k} \]
Thus, the x-intercept is \( A\left(\frac{1}{3k}, 0\right) \).

- **To find the y-intercept (set \( x = 0 \)):**
\[ -2y - 1 = 0 \implies y = -\frac{1}{2} \]
Thus, the y-intercept is \( B\left(0, -\frac{1}{2}\right) \).

**For the line \( 4x - 3y + 2 = 0 \):**

- **To find the x-intercept (set \( y = 0 \)):**
\[ 4x + 2 = 0 \implies x = -\frac{1}{2} \]
Thus, the x-intercept is \( C\left(-\frac{1}{2}, 0\right) \).

- **To find the y-intercept (set \( x = 0 \)):**
\[ -3y + 2 = 0 \implies y = \frac{2}{3} \]
Thus, the y-intercept is \( D\left(0, \frac{2}{3}\right) \).

### Step 2: Use the property of the circle.

The circle passes through points \( A, B, C, \) and \( D \). According to the property of a circle, if a circle passes through four points, the product of the lengths of the segments from the origin to the points on opposite sides must be equal:

\[ OB \cdot OD = OA \cdot OC \]

### Step 3: Calculate the lengths.

- \( OA = \frac{1}{3k} \)
- \( OB = \frac{1}{2} \)
- \( OC = \frac{1}{2} \)
- \( OD = \frac{2}{3} \)

### Step 4: Substitute into the equation.

Substituting the lengths into the equation:

\[ \left(\frac{1}{2}\right) \left(\frac{2}{3}\right) = \left(\frac{1}{3k}\right) \left(\frac{1}{2}\right) \]

### Step 5: Simplify the equation.

Calculating the left side:

\[ \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \]

Calculating the right side:

\[ \frac{1}{3k} \cdot \frac{1}{2} = \frac{1}{6k} \]

Setting the two sides equal:

\[ \frac{1}{3} = \frac{1}{6k} \]

### Step 6: Solve for \( k \).

Cross-multiplying gives:

\[ 1 \cdot 6k = 3 \cdot 1 \implies 6k = 3 \implies k = \frac{3}{6} = \frac{1}{2} \]

Thus, the value of \( k \) is:

\[ \boxed{\frac{1}{2}} \]