A rectangular hyperbola whose centre is C is cut by any circle of radius r in four points P, Q, R and S. Then `CP^(2)+CQ^(2)+CR^(2)+CS^(2)=`

Options

A. `r^(2)`
B. `2r^(2)`
C. `3r^(2)`
D. `4r^(2)`

Answer

To solve the problem, we need to find the sum of the squares of the distances from the center \( C \) of a rectangular hyperbola to the points \( P, Q, R, \) and \( S \) where the hyperbola intersects a circle of radius \( r \).

### Step-by-step Solution:

1. **Understanding the Hyperbola and Circle**:
- The rectangular hyperbola can be represented by the equation \( xy = c^2 \).
- The circle can be represented by the equation \( x^2 + y^2 = r^2 \).
- The center \( C \) of both the hyperbola and circle is at the origin \( (0, 0) \).

2. **Finding Points of Intersection**:
- The points \( P, Q, R, S \) are the intersection points of the hyperbola and the circle.
- To find these points, we substitute \( y = \frac{c^2}{x} \) from the hyperbola equation into the circle equation:
\[ x^2 + \left(\frac{c^2}{x}\right)^2 = r^2 \]
- This simplifies to:
\[ x^2 + \frac{c^4}{x^2} = r^2 \]
- Multiplying through by \( x^2 \) gives:
\[ x^4 - r^2 x^2 + c^4 = 0 \]
- Let \( u = x^2 \). The equation becomes:
\[ u^2 - r^2 u + c^4 = 0 \]

3. **Using Vieta's Formulas**:
- The roots \( u_1, u_2, u_3, u_4 \) of this quadratic equation correspond to \( x_1^2, x_2^2, x_3^2, x_4^2 \) where \( x_i \) are the x-coordinates of points \( P, Q, R, S \).
- By Vieta's formulas, the sum of the roots \( u_1 + u_2 + u_3 + u_4 = r^2 \) and the sum of the products of the roots taken two at a time \( u_1 u_2 + u_1 u_3 + u_1 u_4 + u_2 u_3 + u_2 u_4 + u_3 u_4 = c^4 \).

4. **Calculating Distances**:
- The distances from the center \( C \) to each point \( P, Q, R, S \) are given by:
\[ CP^2 = x_1^2 + y_1^2, \quad CQ^2 = x_2^2 + y_2^2, \quad CR^2 = x_3^2 + y_3^2, \quad CS^2 = x_4^2 + y_4^2 \]
- Since \( y_i = \frac{c^2}{x_i} \), we have:
\[ CP^2 = x_1^2 + \left(\frac{c^2}{x_1}\right)^2 = x_1^2 + \frac{c^4}{x_1^2} \]
- Therefore, the sum of the squares of the distances is:
\[ CP^2 + CQ^2 + CR^2 + CS^2 = \sum_{i=1}^{4} \left(x_i^2 + \frac{c^4}{x_i^2}\right) \]

5. **Final Calculation**:
- Using the results from Vieta's formulas:
\[ \sum_{i=1}^{4} x_i^2 = r^2 \quad \text{and} \quad \sum_{i=1}^{4} \frac{c^4}{x_i^2} = \frac{4c^4}{r^2} \]
- Thus,
\[ CP^2 + CQ^2 + CR^2 + CS^2 = r^2 + \frac{4c^4}{r^2} \]

### Conclusion:
The final result is:
\[ CP^2 + CQ^2 + CR^2 + CS^2 = 4r^2 \]