A line through (4, 2) meets the coordinate axes at P and Q. Then the locus of the circumference of ΔOPQ is 

Concept:

The intercept form of a line is written as follows:

\(\frac{x}{P}+\frac{y}{Q}=1\)

Here, P and Q are the x-intercept and y-intercept respectively.

Calculation:

Let (h, k) is the arbitrary coordinate of the circumcenter of the triangle OPQ.

As it can be observed that two coordinate axes are acting as the two side of the triangle, so the triangle is right triangle and the circumcenter is the midpoint of sides.

\(\begin{align} & \left( h,k \right)=\left( \frac{P+0}{2},\frac{0+Q}{2} \right) \\ & =\left( \frac{P}{2},\frac{Q}{2} \right) \\ \end{align} \)

h = P/2 ⇒ P = 2h

Similarly, k = Q/2 means Q = 2k

The equation of the line \(\frac{x}{P}+\frac{y}{Q}=1\) can be rewritten as:

\(\frac{x}{2h}+\frac{y}{2k}=1\)

Now, this line passes through (4, 2), so it will satisfy the equation of line.

\(\frac{4}{2h}+\frac{2}{2k}=1\)

\(\frac{2}{h}+\frac{1}{k}=1\)

Now to find the locus replace (h, k) with (x, y).

\(\frac{2}{x}+\frac{1}{y}=1\)