If the circle `x^(2)+y^(2)=r^(2)` intersects the hyperbola `xy=c^(2)` in four points `(x_(i),y_(i))` for i=1,2,3 and 4 then `y_(1)+y_(2)+y_(3)+y_(4)=`

Options

A. a) 0
B. b) c
C. c) a
D. d) `c^4`

Answer

To solve the problem, we need to find the sum of the y-coordinates of the intersection points of the circle \( x^2 + y^2 = r^2 \) and the hyperbola \( xy = c^2 \).

### Step-by-step Solution:

1. **Write down the equations:**
We have two equations:
- Circle: \( x^2 + y^2 = r^2 \) (Equation 1)
- Hyperbola: \( xy = c^2 \) (Equation 2)

2. **Express \( x \) in terms of \( y \):**
From Equation 2, we can express \( x \) as:
\[ x = \frac{c^2}{y} \]

3. **Substitute \( x \) into the circle's equation:**
Substitute \( x \) into Equation 1:
\[ \left(\frac{c^2}{y}\right)^2 + y^2 = r^2 \]
Simplifying this gives:
\[ \frac{c^4}{y^2} + y^2 = r^2 \]

4. **Multiply through by \( y^2 \) to eliminate the fraction:**
Multiply the entire equation by \( y^2 \):
\[ c^4 + y^4 = r^2 y^2 \]
Rearranging gives:
\[ y^4 - r^2 y^2 + c^4 = 0 \]

5. **Let \( z = y^2 \):**
Substitute \( z = y^2 \) to transform the equation into a quadratic:
\[ z^2 - r^2 z + c^4 = 0 \]

6. **Use the quadratic formula to find \( z \):**
The roots of the equation are given by:
\[ z = \frac{r^2 \pm \sqrt{r^4 - 4c^4}}{2} \]
Since \( z = y^2 \), we can find \( y \) by taking the square root of \( z \).

7. **Finding the sum of the roots \( y_1 + y_2 + y_3 + y_4 \):**
The equation \( y^4 - r^2 y^2 + c^4 = 0 \) is a quartic equation in \( y \). By Vieta's formulas, the sum of the roots \( y_1 + y_2 + y_3 + y_4 \) can be found by looking at the coefficient of \( y^3 \) (which is 0) divided by the coefficient of \( y^4 \) (which is 1):
\[ y_1 + y_2 + y_3 + y_4 = -\frac{\text{coefficient of } y^3}{\text{coefficient of } y^4} = -\frac{0}{1} = 0 \]

### Final Answer:
Thus, the sum \( y_1 + y_2 + y_3 + y_4 = 0 \).