If $(6, -3)$ is the one extremity of diameter to the circle $x^{2}+y^{2}-3x+8y-4=0$ then its other extremity is-

Given, $(6,-3)$ is one end of diameter.

given, circle equation $x^{2}+y^{2}-3 x+8 y-4=0$

for a general equation, $a x^{2}+b y^{2}+2 h x y+2 g x+2 f y+c=0$, center is $(-g,-f)$

$\therefore$ in $x^{2}+y^{2}-3 x+8 y-4=0$, center will be: $\left(\frac{3}{2},-4\right)$

$\therefore$ Let other end by $(x, y)$;

$\begin{aligned} \therefore \quad \frac{x+6}{2}=\frac{3}{2} \quad \& \quad \frac{y-3}{2}=-4 \\ \Rightarrow \quad x=-3 \quad, \quad y=-5 \end{aligned}$

$\therefore$ the extrenity is $(-3,-5)$

option B is correct.