Step-by-step Solution:

Concept:
The intercept form of a line is written as: \[ \frac{x}{P} + \frac{y}{Q} = 1 \] Here, P and Q are the x-intercept and y-intercept respectively.
Calculation:
Let \((h, k)\) be the arbitrary coordinates of the circumcentre of triangle OPQ.
Since the triangle is right-angled at the origin (O), and the coordinate axes form two sides of the triangle,
the circumcentre will be at the midpoint of the hypotenuse PQ. So, \[ (h, k) = \left( \frac{P + 0}{2}, \frac{0 + Q}{2} \right) = \left( \frac{P}{2}, \frac{Q}{2} \right) \] From this, we get:
\( h = \frac{P}{2} \Rightarrow P = 2h \)
\( k = \frac{Q}{2} \Rightarrow Q = 2k \)
Now substitute into the intercept form of the line: \[ \frac{x}{P} + \frac{y}{Q} = 1 \Rightarrow \frac{x}{2h} + \frac{y}{2k} = 1 \] The line also passes through the fixed point \( (4, 2) \), so it satisfies: \[ \frac{4}{2h} + \frac{2}{2k} = 1 \Rightarrow \frac{2}{h} + \frac{1}{k} = 1 \] Final Step: Locus
Let the circumcentre \( (h, k) \) be represented as a variable point \( (x, y) \).
Then the locus of the circumcentre is given by:
\[ { \frac{2}{x} + \frac{1}{y} = 1 } \]