Explanation

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