A circle cuts the rectangular hyperbola `xy=1` in the points `(x_(r),y_(r)),r=1,2,3,4,`
then `x_(1)x_(2)x_(3)x_(4)=y_(1)y_(2)y_(3)y_(4)=1`.
True or False?

Answer

To determine whether the statement "A circle cuts the rectangular hyperbola \(xy=1\) in the points \((x_{r},y_{r})\) for \(r=1,2,3,4\), then \(x_{1}x_{2}x_{3}x_{4}=y_{1}y_{2}y_{3}y_{4}=1\)" is true or false, we can follow these steps:

### Step 1: Define the Circle and Hyperbola
Let’s consider a circle given by the equation:
\[ x^2 + y^2 = a^2 \]
and the rectangular hyperbola given by:
\[ xy = 1 \]

### Step 2: Parameterize the Hyperbola
We can express points on the hyperbola using a parameter \(t\):
\[ (x, y) = \left(t, \frac{1}{t}\right) \]
where \(t \neq 0\).

### Step 3: Substitute into the Circle's Equation
Substituting the parameterization into the circle’s equation gives:
\[ t^2 + \left(\frac{1}{t}\right)^2 = a^2 \]
This simplifies to:
\[ t^2 + \frac{1}{t^2} = a^2 \]

### Step 4: Clear the Fraction
Multiplying through by \(t^2\) results in:
\[ t^4 - a^2 t^2 + 1 = 0 \]
This is a quadratic equation in terms of \(t^2\).

### Step 5: Solve the Quadratic Equation
Let \(u = t^2\). The equation becomes:
\[ u^2 - a^2 u + 1 = 0 \]
Using the quadratic formula, the roots are:
\[ u = \frac{a^2 \pm \sqrt{a^4 - 4}}{2} \]
Let the roots be \(u_1, u_2, u_3, u_4\) corresponding to \(t_1^2, t_2^2, t_3^2, t_4^2\).

### Step 6: Find the Product of the Roots
From Vieta's formulas, the product of the roots \(u_1 u_2 u_3 u_4\) of the equation \(u^2 - a^2 u + 1 = 0\) is given by:
\[ u_1 u_2 u_3 u_4 = 1 \]
This implies:
\[ t_1^2 t_2^2 t_3^2 t_4^2 = 1 \]

### Step 7: Relate \(x_r\) and \(y_r\) to \(t_r\)
Since \(x_r = t_r\) and \(y_r = \frac{1}{t_r}\), we have:
\[ x_1 x_2 x_3 x_4 = t_1 t_2 t_3 t_4 \]
and
\[ y_1 y_2 y_3 y_4 = \frac{1}{t_1} \cdot \frac{1}{t_2} \cdot \frac{1}{t_3} \cdot \frac{1}{t_4} = \frac{1}{t_1 t_2 t_3 t_4} \]

### Step 8: Calculate the Products
From the earlier step, we know:
\[ t_1 t_2 t_3 t_4 = 1 \implies x_1 x_2 x_3 x_4 = 1 \]
and
\[ y_1 y_2 y_3 y_4 = \frac{1}{1} = 1 \]

### Conclusion
Thus, we conclude that:
\[ x_1 x_2 x_3 x_4 = y_1 y_2 y_3 y_4 = 1 \]
The statement is **True**.