Explanation

(b)
Let the centre of circle on y-axis be (0.k).
Let $d = \sqrt{(7 - 0)^{2} + (3 - k)^{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ d^{2} = 7^{2} + (3 - k)^{2} = D$ (say)
On differentiating w. r. t. k , we get
$\ \ \ \ \ \ \ \ \ \ \ \ \frac{dD}{dk} = 0 + 2(3 - k)( - 1)$
Put $\frac{dD}{dk} = 0 \Rightarrow k = 3$
Now, $\frac{d^{2}D}{dk^{2}} = 2 > 0\ minima,$
$\therefore Minimum$ value at k=3 is d=7
Hence, minimum circle of radius 7.