Explanation

(d)
The two circle whose centre and radius are $C_{1}(0,\ 0),\ r_{1}$=$12,\ C_{2}(3,\ 4),\ r_{2} = 5$ and it passes through origin $ie$, the centre of $C_{1}$
[image:5_image15.jpeg]
Now, $C_{1}C_{2} = \sqrt{3^{2} + 4^{2}} = 5$
And $r_{1} - r_{2} = 12 - 5 = 7$
$\therefore\ \ \ \ \ C_{1}C_{2} < r_{1} - r_{2}$
Hence, circle $C_{2}$ lies inside the circle $C_{1}$
From figure the minimum distance between them, is
$AB = C_{1}B - C_{1}A$
$= r_{1} - (C_{1}C_{2} + C_{2}A)$
$= 12 - 10 = 2$