A line passing through $P(4,2)$ cuts the coordinate axes at $A$ and $B$ respectively. If $O$ is the origin, then the locus of the centre of the circum-circle of $\triangle O A B$ is

• A

  $x^{-1}+y^{-1}=2$
• Correct answer
  B

  $2 x^{-1}+y^{-1}=1$
• C

  $x^{-1}+2 y^{-1}=1$
• D

  $2 x^{-1}+3 y^{-1}=1$

Solution

Let a line cuts the coordinate axes at $A$ and $B$ respectively is

Now, coordinate of centre of the circumcircle of $\triangle O A B$ is mid-point of hypotenuse of right angle $\triangle O A B$ and it is mid-point of $A B$. So, centre of the circumcircle of $\triangle O A B$ is $\left(\frac{a}{2}, \frac{b}{2}\right)$. Now, from Eq. (ii), on taking locus of point $\left(\frac{a}{2}, \frac{b}{2}\right)$, we get $$ \begin{aligned} \frac{2}{x}+\frac{1}{y} & =1 \\ \Rightarrow \quad 2 x^{-1}+y^{-1} & =1 \end{aligned} $$ Hence, option (b) is correct.

Hence, the correct answer is option: 'B'.