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! 

The locus of points with sum of distances $a$ from $(9,0)$ and $(0,9)$ is an ellipse. If we have $a=9\sqrt{2},$ we get a degenerate line segment between the 2 points, but as $a$ increases, the ellipse expands and then becomes tangent to the circle. Thus, you want to find the value of $a$ so that the ellipse with foci at $(9,0)$ and $(0,9)$ is tangent to the circle $x^2+y^2=41.$ Upon finding $a,$ the point of tangency is the desired $z.$
Having completed the interpretation, I leave the calculation to you.
