I have a solution that relies on converting the complex numbers into ordered pairs although I believe there must be a solution with just the help of complex numbers.
Two circles intersect orthogonally, if their radii are perpendicular at the point of intersection. So, using this we can have a condition for orthogonality.
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Here's a trick of how you will get a condition. Let us consider two circles,
$C_1,A:x^2+y^2+2g_1x+2f_1y+c_1=0$
$C_2,B:x^2+y^2+2g_2x+2f_2y+c_2=0$
From your high school course in analytical geometry in high school, it must be clear that the centres $A$ and $B$ are $A(-g_1,-f_1)$ and $B(-g_2,-f_2)$. And the radii of such a circle is $r_1=\sqrt{g_1^2+f_1^2-c_1}$ and similarly $r_2=\sqrt{g_2^2+f_2^2-c_2}$.
Now invoke Pythagoras here, I'll leave the actual computation to you, the condition would turn out to be, $2g_1g_2+2f_1f_2=c_1+c_2$
Now, find a parametric equation for a circle passing through the complex numbers $a$ and $\dfrac{1}{\bar a}$. How do you do this?
Since the circle always passes through, $a\cong(l,m)$ and $\dfrac{1}{\bar a}=\dfrac{a}{|a|^2}\cong\left(\dfrac{l}{l^2+m^2},\dfrac{m}{l^2+m^2}\right)$, you have the following will be the equation of the circle:
$(x-l)\left(x-\dfrac{l}{l^2+m^2}\right)+(y-m)\left(y-\dfrac{m}{l^2+m^2}\right)+\lambda(ly-mx)=0$
The second circle is, $x^2+y^2-1=0$
So, you should now see that $g_2=f_2=0$ and $c_2=-1$. Also, after a little inspection, note that we need not care for what those $g_1$ and $f_1$ are. And, thankfully, $c_1=1$. So, you have the required condition for orthogonality.
I know this is lengthy and not instructive, but this is all I can recollect from high school geometry. So, I only hope this is of some help!