If QR is chord contact when two tangents are drawn from origin to the variable circle, with centre P lying on `xy =1,` if locus of circumcentre is `xy =c^(2)` (where cis positive), then c is `"_____"`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Problem
We are given that QR is the chord of contact when two tangents are drawn from the origin (0, 0) to a variable circle whose center P lies on the hyperbola defined by the equation \(xy = 1\). We need to find the value of \(c\) such that the locus of the circumcenter of triangle OQR is given by \(xy = c^2\).

### Step 2: Define the Coordinates of Point P
Since point P lies on the hyperbola \(xy = 1\), we can express its coordinates as:
- Let \(P(t) = (t, \frac{1}{t})\), where \(t\) is a variable parameter.

### Step 3: Determine the Circumcenter of Triangle OQR
The circumcenter of triangle OQR, where O is the origin (0, 0) and Q and R are points on the circle, can be found as the midpoint of the segment OP.

The coordinates of the circumcenter (C) can be calculated as:
- \(C_x = \frac{0 + t}{2} = \frac{t}{2}\)
- \(C_y = \frac{0 + \frac{1}{t}}{2} = \frac{1}{2t}\)

Thus, the coordinates of the circumcenter are:
\[ C\left(\frac{t}{2}, \frac{1}{2t}\right) \]

### Step 4: Find the Locus of the Circumcenter
To find the locus of the circumcenter, we need to express the product \(C_x \cdot C_y\):
\[ C_x \cdot C_y = \left(\frac{t}{2}\right) \cdot \left(\frac{1}{2t}\right) = \frac{t}{4t} = \frac{1}{4} \]

### Step 5: Relate the Locus to the Given Equation
From the problem, we know that the locus of the circumcenter is given by \(xy = c^2\). We have found that:
\[ xy = \frac{1}{4} \]
Thus, we can equate:
\[ c^2 = \frac{1}{4} \]

### Step 6: Solve for c
Taking the positive square root (since \(c\) is positive):
\[ c = \frac{1}{2} \]

### Final Answer
The value of \(c\) is:
\[ \boxed{\frac{1}{2}} \]