If a circle passes through the point $(a,b)$ and cuts the circle $x^2+y^2=4$ orthogonally, then the locus of its centre is

Let the required circle be $x^2+y^2+2gx+2fy+C=0$.
Its center is $(-g,-f).$ The circle passes through $(a,b):$$a^2+b^2+2ga+2fb+C=0$.Condition for orthogonal intersection with $x^2+y^2=4$ ($x^2+y^2-4=0$):$2g_1{g}_2+2f_1{f}_2=C_1+C_2$ Here $g_2=0,f_2=0,$ and $C_2=-4.$$0+0=C+(-4)\Rightarrow C=4$.
Substitute $C=4$ into the passing condition:
$a^2+b^2+2ga+2fb+4=0$.
To find the locus of the center $(x,y)=(-g,-f),$ substitute $g=-x$ and $f=-y$:
$a^2+b^2+2(-x)a+2(-y)b+4=0$
$a^2+b^2-2ax-2by+4=0$
Rearranging:
$2ax+2by-(a^2+b^2+4)=0$.