Explanation

(b)
Equation of line is
$\frac{x}{a} + \frac{y}{b} = 1$ …(i)
Let $P$ be the foot of the perpendicular drawn from the origin to the line whose coordinates are $(x_{1},y_{1})$
Since, $OP\bot AB$
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$\therefore$Slope of$\ OP \times$Slope of$\ AB = - 1$
$\Rightarrow \ \left( \frac{y_{1}}{x_{1}} \right)\left( \frac{b}{- a} \right) = - 1$
$\Rightarrow by_{1} = ax_{1}$ …(ii)
Since, $P$ lies on the line $AB$, then
$\frac{x_{1}}{a} + \frac{y_{1}}{b} = 1$
$\Rightarrow bx_{1} + ay_{1} = ab$ …(iii)
From Eq. (ii) and (iii), we get
$x_{1} = \frac{ab^{2}}{a^{2} + b^{2}}\ \ and\ \ y_{1} = \frac{a^{2}b}{a^{2} + b^{2}}$
Now, $x_{1}^{2} + y_{1}^{2} = \left( \frac{ab^{2}}{a^{2} + b^{2}} \right)^{2} + \left( \frac{a^{2}b}{a^{2} + b^{2}} \right)^{2}$
$\Rightarrow x_{1}^{2} + y_{1}^{2} = \frac{a^{2}b^{4}}{\left( a^{2} + b^{2} \right)^{2}} + \frac{a^{4}b^{2}}{\left( a^{2} + b^{2} \right)^{2}}$
$\Rightarrow x_{1}^{2} + y_{1}^{2} = \frac{a^{2}b^{2}(a^{2} + b^{2})}{\left( a^{2} + b^{2} \right)^{2}}$
$\Rightarrow x_{1}^{2} + y_{1}^{2} = \frac{a^{2}b^{2}}{(a^{2} + b^{2})}$
$\Rightarrow x_{1}^{2} + y_{1}^{2} = \frac{1}{\frac{1}{a^{2}} + \frac{1}{b^{2}}}$
But $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}$ (given)
$\therefore\ \ x_{1}^{2} + y_{1}^{2} = c^{2}$
Thus, the locus of $P(x_{1},y_{1})$ is
$x^{2} + y^{2} = c^{2}$
Which is the equation of circle