For all real values of k, the polar of the point (2k, k-4) with respect to `x^(2) + y^(2) - 4x - 6y + 1 = 0` passes through the point

Options

A. (1,1)
B. (1,-1)
C. (-3,1)
D. (3,1)

Answer

To solve the problem step by step, we need to find the polar of the point (2k, k-4) with respect to the given circle and determine the point through which this polar passes for all real values of k.

### Step 1: Write the equation of the circle
The equation of the circle is given as:
\[ x^2 + y^2 - 4x - 6y + 1 = 0 \]

### Step 2: Identify the coefficients
From the equation of the circle, we can identify the coefficients:
- \( g = -4 \)
- \( f = -6 \)
- \( c = 1 \)

### Step 3: Find the polar of the point (2k, k-4)
The polar of a point \((x_1, y_1)\) with respect to the circle is given by:
\[ x \cdot x_1 + y \cdot y_1 + g \cdot x + f \cdot y + c = 0 \]

Substituting \( x_1 = 2k \) and \( y_1 = k - 4 \):
\[ x \cdot (2k) + y \cdot (k - 4) - 4x - 6y + 1 = 0 \]

### Step 4: Expand the equation
Expanding the equation gives:
\[ 2kx + ky - 4y - 4x - 6y + 1 = 0 \]
This simplifies to:
\[ 2kx + ky - 4x - 10y + 1 = 0 \]

### Step 5: Rearrange the equation
Rearranging the terms, we have:
\[ k(2x + y) + (-4x - 10y + 1) = 0 \]

### Step 6: Set conditions for the equation to be independent of k
For this equation to hold for all real values of \( k \), the coefficient of \( k \) must be zero:
\[ 2x + y = 0 \quad \text{(1)} \]

And the constant term must also equal zero:
\[ -4x - 10y + 1 = 0 \quad \text{(2)} \]

### Step 7: Solve the system of equations
From equation (1):
\[ y = -2x \]

Substituting \( y = -2x \) into equation (2):
\[ -4x - 10(-2x) + 1 = 0 \]
This simplifies to:
\[ -4x + 20x + 1 = 0 \implies 16x + 1 = 0 \implies x = -\frac{1}{16} \]

Now substituting \( x = -\frac{1}{16} \) back into \( y = -2x \):
\[ y = -2\left(-\frac{1}{16}\right) = \frac{1}{8} \]

### Step 8: Write the final coordinates
Thus, the polar of the point passes through the point:
\[ \left(-\frac{1}{16}, \frac{1}{8}\right) \]

### Step 9: Conclusion
The polar of the point (2k, k-4) with respect to the given circle passes through the point \(\left(-\frac{1}{16}, \frac{1}{8}\right)\).