If tangents `O Q` and `O R` are dawn to variable circles having radius `r` and the center lying on the rectangular hyperbola `x y=1` , then the locus of the circumcenter of triangle `O Q R` is `(O` being the origin)
(a). `x y=4` (b) `x y=1/4` `x y=1` (d) none of these

Options

A. xy = 4
B. xy = 1/4
C. xy = 1
D. none of these

Explanation

To solve the problem, we need to determine the locus of the circumcenter of triangle \( OQR \) where \( O \) is the origin, and \( Q \) and \( R \) are points on the tangents drawn from the origin to circles centered on the rectangular hyperbola \( xy = 1 \).

### Step 1: Identify the center of the circles
The centers of the circles lie on the rectangular hyperbola \( xy = 1 \). A general point on this hyperbola can be expressed as \( (t, \frac{1}{t}) \).

### Step 2: Define the radius of the circles
Let the radius of the circles be \( r \). The tangents from the origin \( O(0, 0) \) to a circle centered at \( C(t, \frac{1}{t}) \) with radius \( r \) can be determined using the formula for the length of the tangent from a point to a circle.

### Step 3: Equation of the tangents
The equation of the tangent from the origin to the circle centered at \( C(t, \frac{1}{t}) \) can be derived. The distance \( d \) from the origin to the center \( C(t, \frac{1}{t}) \) is given by:
\[ d = \sqrt{t^2 + \left(\frac{1}{t}\right)^2} = \sqrt{t^2 + \frac{1}{t^2}} = \sqrt{\frac{t^4 + 1}{t^2}} = \frac{\sqrt{t^4 + 1}}{t} \]
The length of the tangent \( L \) from the origin to the circle is given by:
\[ L = \sqrt{d^2 - r^2} = \sqrt{\left(\frac{\sqrt{t^4 + 1}}{t}\right)^2 - r^2} = \sqrt{\frac{t^4 + 1}{t^2} - r^2} \]

### Step 4: Finding the coordinates of points \( Q \) and \( R \)
The points \( Q \) and \( R \) lie on the tangents drawn from the origin to the circle. The coordinates of these points can be expressed in terms of the angle \( \theta \) that the tangent makes with the x-axis.

### Step 5: Determine the circumcenter of triangle \( OQR \)
The circumcenter of triangle \( OQR \) is the midpoint of the segment joining the origin \( O(0, 0) \) and the center \( C(t, \frac{1}{t}) \). Thus, the coordinates of the circumcenter \( M \) can be expressed as:
\[ M\left(\frac{t}{2}, \frac{1/t}{2}\right) = \left(\frac{t}{2}, \frac{1}{2t}\right) \]

### Step 6: Eliminate \( t \) to find the locus
Let \( x = \frac{t}{2} \) and \( y = \frac{1}{2t} \). From \( x \), we can express \( t \) as:
\[ t = 2x \]
Substituting this into the equation for \( y \):
\[ y = \frac{1}{2(2x)} = \frac{1}{4x} \]
Multiplying both sides by \( 4x \) gives:
\[ xy = \frac{1}{4} \]
Thus, the locus of the circumcenter is given by the equation:
\[ xy = \frac{1}{4} \]

### Conclusion
The correct answer is (b) \( xy = \frac{1}{4} \).