The locus of a point which moves such that the difference of the squares of lengths of tangents drawn from it to two given circles is constant, is :

Options

A. a circle having radius equal to sum of radii of given circles
B. A line perpendicular to the line joining the centres of the circle
C. A line parallel to the direct common tangents of the circle
D. A circle having radius `sqrt(r_(1)r_(2)),r_(1),r_(2)` being the radii of given circles

Explanation

To solve the problem, we need to find the locus of a point \( P(h, k) \) such that the difference of the squares of the lengths of the tangents drawn from this point to two given circles is constant.

### Step-by-Step Solution:

1. **Equations of the Circles**:
Let the equations of the two circles be:
\[ C_1: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 \]
\[ C_2: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \]

2. **Length of Tangents**:
The length of the tangent from a point \( P(h, k) \) to a circle defined by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:
\[ L = \sqrt{h^2 + k^2 + 2gh + 2fk + c} \]

For circle \( C_1 \), the length of the tangent \( L_1 \) is:
\[ L_1 = \sqrt{h^2 + k^2 + 2g_1h + 2f_1k + c_1} \]
For circle \( C_2 \), the length of the tangent \( L_2 \) is:
\[ L_2 = \sqrt{h^2 + k^2 + 2g_2h + 2f_2k + c_2} \]

3. **Difference of Squares of Lengths**:
We are given that the difference of the squares of the lengths of the tangents is constant. Thus, we can write:
\[ L_1^2 - L_2^2 = k \]
where \( k \) is a constant.

4. **Expanding the Squares**:
Expanding \( L_1^2 \) and \( L_2^2 \):
\[ L_1^2 = h^2 + k^2 + 2g_1h + 2f_1k + c_1 \]
\[ L_2^2 = h^2 + k^2 + 2g_2h + 2f_2k + c_2 \]

5. **Setting Up the Equation**:
Now substituting into the difference:
\[ (h^2 + k^2 + 2g_1h + 2f_1k + c_1) - (h^2 + k^2 + 2g_2h + 2f_2k + c_2) = k \]
Simplifying this gives:
\[ 2g_1h + 2f_1k + c_1 - 2g_2h - 2f_2k - c_2 = k \]

6. **Rearranging Terms**:
Rearranging the terms leads to:
\[ 2(g_1 - g_2)h + 2(f_1 - f_2)k + (c_1 - c_2 - k) = 0 \]

7. **Identifying the Locus**:
This equation represents a linear equation in \( h \) and \( k \), which can be rewritten as:
\[ (g_1 - g_2)h + (f_1 - f_2)k + \frac{1}{2}(c_1 - c_2 - k) = 0 \]
This is the equation of a straight line.

### Conclusion:
Thus, the locus of the point \( P(h, k) \) is a straight line.