Explanation

(d)
Since, line $OA$ makes an angle $\alpha$ with $x$-axis and given $OA = a,$ then coordinates of $A$ are $(a\cos\alpha,\ a\sin{\alpha)}$. Also, $OB\bot OA$, then $OB$ makes an angle $(90{^\circ} + \alpha)$ with $x$-axis, then coordinates of $B$ are $\lbrack a\cos(90{^\circ} + \alpha),\ a\sin{(90{^\circ} + \alpha)\rbrack\ }$
$ie,\ \left( - a\sin\alpha,\ a\cos\alpha \right)$
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Equation of the diagonal $AB$ not passing through the origin is
$(y - a\sin\alpha) = \frac{a\cos\alpha - a\sin\alpha}{- a\sin{\alpha - a\cos\alpha}}(x - a\cos{\alpha)}$
$\Rightarrow \ (\sin{\alpha + \cos{\alpha)(y - a\sin{\alpha) = (\sin{\alpha - \cos{\alpha)(x - a\cos{\alpha)}}}}}}$
$\Rightarrow \ y\left( \sin\alpha + \cos\alpha \right) + x\left( \cos\alpha - \sin\alpha \right)$
$= a\sin\alpha\left( \sin\alpha + \cos\alpha \right) - a\cos\alpha\left( \sin\alpha - \cos\alpha \right)$
$= a\ \left( \sin^{2}\alpha + \sin\alpha\cos\alpha - \cos\alpha\sin\alpha + \cos^{2}\alpha \right)$
$\Rightarrow \ y\left( \sin\alpha + \cos\alpha \right) + x\left( \cos\alpha - \sin\alpha \right) = a\$