Explanation

(b)
We have, $\sqrt{(x - 3)^{2} + (y - 1)^{2}} + \sqrt{(x + 3)^{2} + (y - 1)^{2}} = 6$
$\Rightarrow \ \sqrt{(x - 3)^{2} + (y - 1)^{2}} = 6 - \sqrt{(x + 3)^{2} + (y - 1)^{2}}$
On squaring both sides, we get
$12x + 36 = 12\sqrt{(x + 3)^{2} + ({y - 1)}^{2}}$
$\Rightarrow x + 3 = \sqrt{(x + 3)^{2} + ({y - 1)}^{2}}$
Again on squaring, we get
$x^{2} + 9 + 6x = x^{2} + 9 + 6x + y^{2} + 1 - 2y$
$\Rightarrow \ y^{2} - 2y + 1 = 0$
Which represents a pair of straight lines