Equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-4x-6y-3=0$  is:

Equation of the circle through origin is

$x^2+y^2+2gx+2fy=0$                                              ……(1)

It intersects the circles

$x^2+y^2-8y+12=0$  and $x^2+y^2-4x-6y-3=0$  orthogonally, so

$0-8f=12+0\Rightarrow f=-3/2$

and    $-4g-6f=-3+0\Rightarrow g=3$ .

Hence, the required circle is

$x^2+y^2+6x-3y=0$ .