Well, since the point $D$ is definitively not on $PQ$, we will "slightly change" the statement, so that the expected answer is an answer, according to the following picture:
Problem:
$\Delta PQR$ is an equilateral triangle with points in the cartesian plane $\Bbb R^2$, so that the equation of the line $PQ$ is $\sqrt 3 x +y-6=0$. The incircle $\odot (I)$ of $\Delta PQR$ touches the sides of this triangle opposite to $P,Q,R$ in $D,E,F$ respectively.
The point $D$ has coordinates $\displaystyle\left(\frac {\sqrt 3}2,\frac 32\right)$.
Find the equation of the incircle $\odot(I)=\odot(DEF)$.
Note: I could not understand the "same side" story.
Note:
I am against formulations of the shape <<If $PQ$ is ... and $D$ is ... and ... find equation of the circle $\odot(I)$>>
Why? Since this "if-then sentence" is not giving anything, it is an implication. Even if we understand the if-condition, and would be polite to assume it, it would be false to solve the problem till we can show that the assumption is true. But we do not have enough information to decide. I would not recommend to do this in an exam, as i did, except for the case of an ultima ratio. Well, i had luck, the professor accepted my logic.
Solution:
Since $\Delta PQR$ is equilateral, the triangle $\Delta DEF$ is also equilateral.
Its circumcircle $\odot(I)$ has radius $1$, so the two sides of equal length $1$ in $\Delta IDF$ are building an angle of $120^\circ$, which gives
$$
DF^2=1^2+1^2-2\cdot 1\cdot 1\cdot \cos 120^\circ
=1+1+1=3\ .
$$
So $F\in PQ$ is a point on the line $PQ$ with known equation, at known distance $\sqrt 3$ from the known point $D$. Very quickly the "two chances for $F$", let us denote them by $F_1$ and $F_2$ are
$$
F_1 =
\left(
\frac{3\sqrt 3}2,\frac 32
\right) \ ,
\qquad
F_2=(\sqrt 3,3)\ .
$$
We consider the two chances separately.
- For $F=F_1$, since the point $Q$ makes $\Delta DFQ$ equilateral, and $Q$ satisfies the given equation of $PQ$, we have only the chance $Q=F_2$. The homothety centered in $Q$ and factor $2$ brings $\Delta QDF$ into $\Delta QRP$, so the other two vertices of the latter are $R(0,0)$ and $P(2\sqrt 3,0)$. The mid point of $PR$ is thus $E=(\sqrt 3,0)$. (Or we could compute $E$ as the reflection of $Q$ in $DF$.) The median $QE$ between $Q(\sqrt3,3)$ and $E(\sqrt 3,0)$ has length $3$, and contains the point $I$, which is also the centroid, placed at distances in proportion $2:1$ from $Q$ and $E$, so
$$
I=(\sqrt 3,1)\ .
$$
(Well, this point and the origin both satisfy $\sqrt 3 x +y-6<0$, so they are on the same side of $PQ$, but the next point has the same property.)
It remains to write down the equation of the circle centered in this point $I$ having radius one, $(x-\sqrt 3)^2+(y-1)^2=1^2$.
- For $F=F_2$, with the same argument, $Q$ is the other "$F$-choice", $Q=F_1$, and we construct similarly $I$. The reflection of $Q$ w.r.t. $DF$ is the point $E(0,3)$, so the second choice of $I$ is $\displaystyle\left(\frac{\sqrt 3}2,\frac 52\right)$.
