Intersection of circles toward the left of the blue line

Consider a straight line of length $1$ unit on the $y-\text{axis}$. Name it $AF$. Take another point $D$ below the $x-\text {axis}$ such that it has a negative $y-\text {coordinate}$ and a positive $x-\text {coordinate}$.

Show that the circle with radius $DF$ and the circle with radius $AF$ will always meet at a point towards the left of the left of the line $DA$. (In other words, if the equation of the blue line is $y=-mx+c$ then show that the intersection point will be at a point such that $y<mx+c$)

I'm sorry I cannot put any of my work because I couldn't do it. Perhaps a coordinate solution will be okay (and I could do that) but I am interested in a pure Euclidean geometry solution.

• Infection of circles?! Is this pathological geometry? Mar 29, 2018 at 13:23
• @Théophile Oops, edited. Mar 29, 2018 at 13:28
• Is it okay now? Mar 29, 2018 at 14:11
• Yes, that makes more sense. :) Mar 29, 2018 at 14:24

Both circles are symmetric about line $AD$, so if they intersect at $F$ they will also intersect at the reflection of $F$ about $AD$.