A line passing through P(4,2) cuts the coordi | vedclass

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(B) Let the equation of the line cutting the axes at $A(a, 0)$ and $B(0, b)$ be $\frac{x}{a} + \frac{y}{b} = 1$. Since the line passes through $P(4, 2

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

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(B) Let the equation of the line cutting the axes at $A(a, 0)$ and $B(0, b)$ be $\frac{x}{a} + \frac{y}{b} = 1$. Since the line passes through $P(4, 2)$,we have $\frac{4}{a} + \frac{2}{b} = 1$ ... $(i)$. In the right-angled $	riangle OAB$,the circum-centre is the midpoint of the hypotenuse $AB$. Let the circum-centre be $(h, k)$. Then $h = \frac{a}{2} \Rightarrow a = 2h$ and $k = \frac{b}{2} \Rightarrow b = 2k$. Substituting these into equation $(i)$,we get $\frac{4}{2h} + \frac{2}{2k} = 1$,which simplifies to $\frac{2}{h} + \frac{1}{k} = 1$. Replacing $(h, k)$ with $(x, y)$,the locus is $2x^{-1} + y^{-1} = 1$. Thus,option $(b)$ is correct.

$A$ 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 $\triangle OAB$ is

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