Line $p$ and circles $k$ and $l$ are given. Construct point $T$ on line $p$ such that tangents to circles $k$ and $l$ from point $T$ form equal angles with line $p$. How many solutions are there?
I don't know why but I just drew a line $t$ parallel to $p$ passing through the center of circle $l$. Then I found the perpendicular bisector of $IJ$ where $I=t\cap l$ and $J=t \cap k$. I thought the interesction of that perpendicular bisector with the line $p$ would be point $T$. But apparently it's not even tho it's close. Any hints would be appreciated.
What I'm trying to achieve:
As shown in figure, mirror circle d about the line to get circle d'. Draw common tangent i of circles d' and c, it intersect the line at P. Draw a tangent j from P on circle d. Tangents i and j are solutions. Now draw another common tangent k passing point S on circle c, it intersect the line at T. Draw tangent l from T on the circle d passing point U. Tangent l and TS are another solution. Could you find the reason?