A Circle Problem A circle $K$, a tangent $T$ and a point $A$ on $t$ are given. Find the locus of all point $X$ for which points $Y$ and $Z$ on $T$ exist which are equidistant from $A$ and make $K$ the incircle of the triangle $XYZ.$
I suppose that the locus of all point $X$ is a straight but I can't prove it.
I hope you can give me suggestions or hints for my problem.
Thanks.
 A: Let $B$ be the tangency point, and let $B'$ be a point symmetric to this point in point $A$. Also let $P$ be a point antipodal to $B$ on $K$. Now $B'$ is a point of tangency of excircle of triangle $XYZ$ opposite to vertex $X$. Now it's easy to see that a homothety taking the incircle to excircle with center at $X$ moves $P$ to $B'$, which means that $X,P,B'$ are collinear, so $X$ lies on line $PB'$. You should also be able to show that in order for $XYZ$ to have K as an incircle we need $X$ to lie on an open ray contained in line $PB'$ with one end at $P$ and not containing $B'$ (in some other cases we can make K be an excircle, but not an incircle).
To show that every point $X$ on this ray satisfies the requirement, take two tangents to K passing through $X$ and call intersections of these tangents with line t $Y$ and $Z$. By using arguments similar to above we can show that $Y,Z$ are equidistant from $A$.

Edit: here is why $P$ moves to $B'$: Let me denote the excircle by E here. Note that the line tangent to K at $P$ is parallel to line t. Because image of line under a homothety is parallel to original line, we get that image if this tangent is parallel (or equal to) line t. But this image will also be tangent to E (because E is the image of K). I hope it's clear enough why image of this tangent to K must then be t. Lastly, the image of tangency point is a tangency point, so image of $P$ will be tangency point of E to t, which is precisely $B'$.
