Find all complex numbers satisfying $z\cdot\bar{z}=41$, for which $|z-9|+|z-9i|$ has the minimum value My first attempt was to express $z$ as $x+iy$ and minimize the expression $\sqrt{(x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}$ where $x^2+y^2=41$. 
That said, it seems to me that using the geometric interpretation could be easier. As far as I understand, I need to find points on the circle for which the sum of distances to the points $(9,0)$ and $(0,9)$ is lowest. This interpretation, however, doesn't help with regard to calculations. 
Is there some simple trick or idea I'm missing?
Thank you! 

Hint: Show that $$\sqrt{(8x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}\geq 9\sqrt{2}$$ and the equal sign holds if $$x=4,y=5$$
ok we will prove the inequality above:
squaring all we get
$$2\sqrt{(8x-9)^2+y^2}\sqrt{(x^2+(y-9)^2}\geq 162-(8x-9)^2-(y-9)^2-x^2-y^2$$,
now we use that $x^2+y^2=41$:
we get
$$2\sqrt{(8x-9)^2+41-x^2}\sqrt{41-y^2+(y-9)^2}\geq 121-(8x-9)^2-(y-9)^2$$
squaring again and factorizing we get
$$- \left( 64\,{x}^{2}+{y}^{2}-144\,x-18\,y+41 \right)  \left( 64\,{x}^{
2}+{y}^{2}-144\,x-18\,y+42 \right) 
\geq 0$$
which is true.
