A circle $S$ passes through the point $(0, 1)$ and is orthogonal to the circles $(x-1)^{2}+y^{2}=16$ and $x^{2}+y^{2}=1$. Then (1) radius of $S$ is $8$ (2) radius of $S$ is $7$ (3) centre of $S$ is $(-7, 1)$ (4) centre of $S$ is $(-8, 1)$

$\text{Given: } S \text{ passes through } (0, 1)$

$S \perp (x-1)^2 + y^2 = 16$

$S \perp x^2 + y^2 = 1$

$\text{To find: which option gives the radius / centre of } S$

$S: x^2 + y^2 + 2gx + 2fy + c = 0$

$\text{Orthogonal: } 2(g_1 g_2 + f_1 f_2) = c_1 + c_2$

$C_1: x^2 + y^2 - 2x - 15 = 0 \Rightarrow g_1 = -1, f_1 = 0, c_1 = -15$

$C_2: x^2 + y^2 - 1 = 0 \Rightarrow g_2 = 0, f_2 = 0, c_2 = -1$

$2(g\cdot 0 + f\cdot 0) = c + (-1)$

$\Rightarrow c = 1$

$2(g(-1) + f\cdot 0) = c + (-15) \Rightarrow -2g = c - 15$

$c = 1 \Rightarrow -2g = -14 \Rightarrow g = 7$

$\text{Put } (0,1):\ 1 + 2f + c = 0$

$c = 1 \Rightarrow 2f = -2 \Rightarrow f = -1$

$S: x^2 + y^2 + 14x - 2y + 1 = 0$

$\text{Centre} = (-g, -f) = (-7, 1)$

$r = \sqrt{g^2 + f^2 - c}$

$r = \sqrt{49 + 1 - 1} = \sqrt{49} = 7$

$d^2 = (1+7)^2 + (0-1)^2 = 64 + 1 = 65$

$r^2 + r_1^2 = 7^2 + 4^2 = 49 + 16 = 65 \ \checkmark$

$\text{Centre} = (-7, 1) \Rightarrow \text{option (3)}$

$r = 7 \Rightarrow \text{option (2)}$

$\boxed{\text{Options (2) and (3) are correct}}$