The locus of the foot of the perpendicular from the centre of the hyperbola `xy = c^2` on a variable tangent is (A) `(x^2-y^2)=4c^2xy` (B) `(x^2+y^2)^2=2c^2xy` (C) `(x^2+y^2)=4c^2xy` (D) `(x^2+y^2)^2=4c^2xy`

Options

A. `(x^(2)-y^(2))^(2) = 4c^(2)xy`
B. `(x^(2)+y^(2))^(2) =2c^(2)xy`
C. `(x^(2)-y^(2))^(2)=2c^(2)xy`
D. `(x^(2)+y^(2))^(2)=4c^(2)xy`

Explanation

To find the locus of the foot of the perpendicular from the center of the hyperbola \( xy = c^2 \) on a variable tangent, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Hyperbola**: The given hyperbola is \( xy = c^2 \). This is a rectangular hyperbola centered at the origin (0,0).

2. **Parametric Representation**: The points on the hyperbola can be represented parametrically as \( P(ct, \frac{c}{t}) \), where \( t \) is a parameter.

3. **Equation of the Tangent**: The equation of the tangent to the hyperbola at point \( P \) can be derived using the parametric form:
\[ \frac{x}{ct} + \frac{y}{\frac{c}{t}} = 2 \]
Simplifying this gives:
\[ x + yt^2 = 2ct \]

4. **Slope of the Tangent**: The slope of the tangent line can be found from the equation:
\[ \text{slope} = -\frac{1}{t^2} \]

5. **Slope of the Perpendicular**: Since the foot of the perpendicular from the center (0,0) to the tangent line has a slope that is the negative reciprocal of the tangent slope, we have:
\[ \text{slope of CM} = t^2 \]

6. **Equation of the Perpendicular**: The equation of the line from the center (0,0) with slope \( t^2 \) is:
\[ y = t^2 x \]

7. **Finding the Coordinates of the Foot of the Perpendicular**: Let the foot of the perpendicular be \( M(h, k) \). Since \( M \) lies on the line \( y = t^2 x \), we can express:
\[ k = t^2 h \]

8. **Substituting into the Tangent Equation**: The point \( M(h, k) \) also lies on the tangent line, so substituting \( k = t^2 h \) into the tangent equation gives:
\[ h + t^2(t^2 h) = 2ct \]
Simplifying this gives:
\[ h + t^4 h = 2ct \]
or
\[ h(1 + t^4) = 2ct \]
Thus,
\[ h = \frac{2ct}{1 + t^4} \]
and substituting back to find \( k \):
\[ k = t^2 h = \frac{2ct^3}{1 + t^4} \]

9. **Finding the Locus**: Now we need to eliminate the parameter \( t \). We have:
\[ h = \frac{2ct}{1 + t^4}, \quad k = \frac{2ct^3}{1 + t^4} \]
From \( h \), we can express \( t \):
\[ t = \frac{h(1 + t^4)}{2c} \]
Substituting this expression for \( t \) into the equation for \( k \) and simplifying will lead to a relation involving \( h \) and \( k \).

10. **Final Equation**: After substituting and simplifying, we arrive at:
\[ (x^2 + y^2)^2 = 4c^2xy \]

### Conclusion:
The locus of the foot of the perpendicular from the center of the hyperbola \( xy = c^2 \) on a variable tangent is given by the equation:
\[ (x^2 + y^2)^2 = 4c^2xy \]
Thus, the correct option is **(D)**.