Question

A line passing through the point $P(4,2)$ , meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. If $O$ is the origin, then locus of the center of the circumcircle of $△OAB$ is

• A. $x^{-1}+y^{-1}=2$
• B. $2x^{-1}+y^{-1}=2$
• C. $x^{-1}+2y^{-1}=1$
• D. $2x^{-1}+2y^{-1}=1$

Answer 1

The correct option is B $2x^{-1}+y^{-1}=2$

Let the coordinates of $A$ and $B$ be $(a,0)$ and $(0,b)$ respectively.

Then, equation of line $AB$ is

$\frac{x}{a}+\frac{y}{b}=1$

Since, it passes through the point $P(4,2)$

$\therefore \frac{4}{a}+\frac{2}{b}=1$ ...$\left(1\right)$

Now, center of the circumcircle of $△OAB=(\frac{a}{2},\frac{b}{2})$

So, equation (1) can be written in the form $\frac{2}{\left(\frac{a}{2}\right)}+\frac{1}{\left(\frac{b}{2}\right)}=1$

$\therefore$ Locus of circumcenter is

$\frac{2}{x}+\frac{1}{y}=1\Rightarrow 2x^{-1}+y^{-1}=2$