The center of the pink circle is at the intersection of a vertical at distance $r$ ($35/2$ mm) of the pink line, and a circle of radius $2r$ concentric with the blue circle.
To find this intersection numerically, solve
$$(x-x_c)^2+(y-y_c)^2=4r^2,\\y=y_v,$$$$(x-x_c)^2+(y-y_c)^2=4r^2,\\x=x_v+r,$$ which is trivial.
The center of the pink circle is at the intersection of a vertical at distance $r$ ($35/2$ mm) of the pink line, and a circle of radius $2r$ concentric with the blue circle.
To find this intersection, solve
$$(x-x_c)^2+(y-y_c)^2=4r^2,\\y=y_v,$$ which is trivial.
The center of the pink circle is at the intersection of a vertical at distance $r$ ($35/2$ mm) of the pink line, and a circle of radius $2r$ concentric with the blue circle.
To find this intersection numerically, solve
$$(x-x_c)^2+(y-y_c)^2=4r^2,\\x=x_v+r,$$ which is trivial.
