Given equation of circle is $x^2+y^2-10 x+12 y-3=0$ $\therefore g=-5, f=6, c=-3$ centre $=(5,-6)$ and $r=8$ We know that that pole of polar $A x+B y+C=0$ w.r to circle $x^2+y^2+2 g x+2 f y+c=0$ can be find out by $\frac{x_1+g}{A}=\frac{y_1+f}{B}=\frac{g x_1+t y_1+c}{c}$
from option (d) $A=6, B=-8, C=15$
$\therefore \frac{x_1-5}{6}=\frac{y_1+6}{-8}=\frac{-5 x_1+6 y_1-3}{15}$
$\Rightarrow 15 x_1-12 y_1=19... (i)$
and $40 x_1-63 y_1=66... (ii)$
From equations (i) and (ii), we get
$x_1=\frac{135}{155}, \quad y_1=\frac{-46}{93}$
Now, distance of pole from centre
$\left(\frac{135}{155}-5\right)^2+\left(\frac{-46}{93}+6\right)^2=47.30 < 8^2 \text { (radius) }$
So, pole inside the circle that shows polar is not a tangent and not a chord of contact.