For the following question show that locus is a circle. Two circles touch one another internally at A, and a variable chord PQ of the outer circle touches the inner circle. Show that the locus of the incentres of the triangle APQ is another circle touching the given circles at A.
First I found the equation of the chord which is also the tangent to the smaller circle. Then I thought a lot on how to continue further;  but literally have no idea how to do it. Even if we use the formula of the incetre of a triangle then we have to find the length of the sides of the triangle which is damn tedious.
Please help me with shorter method.
 A: Let $H$, $K$ be the points where chords $AP$, $AQ$ intersect the inner circle (see figure). The homothety of centre $A$ carrying the outer circle to the inner circle also carries $P$ to $H$ and $Q$ to $K$, with
$$
{AP\over AH}={AQ\over AK}={R\over r},
$$
where $r$ and $R$ are the radii of inner and outer circle respectively.
Hence $HK$ is parallel to tangent $PQ$ and tangency point $T$ divides arc $HK$ into two equal parts.It follows that inscribed angles $\angle HAT$ and $\angle KAT$ are equal too and $AT$ is the bisector of $\angle PAQ$, so that incenter $I$ lies on $AT$.
Moreover, by the angle bisector theorem we have $IT:IA=PT:PA$, that is:
$$
{IT^2\over IA^2}={PT^2\over PA^2}={PH\cdot PA\over PA^2}=
{PH\over PA}={PA-HA\over PA}=1-{HA\over PA}=1-{r\over R}.
$$
Hence $AT/AI$ is a fixed ratio:
$$
{AT\over AI}={AI+IT\over AI}=1+{IT\over AI}=1+\sqrt{1-{r\over R}}
$$
and incenter $I$ is the image of $T$ under the homothety of centre $A$ and ratio given above. Hence its locus is a circle, the image of the inner circle under the same homothety.

A: Given two circles internally tangent at $A$, and any chord $PQ$ in the greater circle tangent to the lesser circle, prove that the incenter of triangle $APQ$ lies on a circle which is also internally tangent at $A$.
$B$ is center of the greater circle, $AC$ the diameter of the lesser, and $D$ the point of tangency.
Join $AD$ and extend it to $F$. Join $DC$ and $FB$ intersecting at $J$.
Let $QG$ bisect $\angle AQP$, and draw $GH\parallel DC$.

Since$$\angle JDE=\angle CAD=\angle JFD$$and complements$$\angle DJF=\angle DCA$$then$$\triangle ADC\sim\triangle FDJ\sim\triangle DEJ$$Therefore $\angle JED$ is also right, making $QE=EP$, arc $QF$=arc $PF$, and$$\angle QAF=\angle PAF$$
Hence $G$ is the incenter of $\triangle APQ$.
And since by angle bisection and parallels$$\frac{QA}{QD}=\frac{GA}{GD}=\frac{HA}{HC}$$and point $C$ is fixed, then $H$ is also fixed.
And since $\angle AGH$ is right, it follows that incenter $G$ lies on the circumference of a circle with diameter $AH$ and internally tangent at $A$.
