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 centre of the circum circle of $\Delta OAB$ is

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

Answer 1

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

Let the x-intercept be $(a,0)$ and y-intercept be $(0,b)$.

As $△AOB$ is a right angled triangle, it's hypotenuse will bethe diametre of the circum circle.

$\because$ diametre of a circle subtends a right angle at the arc of the circle.

Therefore the circumcentre will be $(\frac{a}{2},\frac{b}{2})$ (Section Formula)

As the lines pass through $(4,2)$,

the equation of all those lines can be written as: $(y-2)=m(x-4)$, where m is the slope of the line.

Putting points $(a,0)$ and $(b,0)$ in the equation, we get,

$a=\frac{4m-2}{m}$ and $b=2-4m$

ie, $\frac{4}{a}+\frac{2}{b}=\frac{2m}{2m-1}-\frac{1}{2m-1}=1$

Inorder to fing the locus of all points $(\frac{a}{2},\frac{b}{2})$,

Substitute $x=\frac{a}{2}$ and $y=\frac{b}{2}$

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

Therefore option B is the correct answer.