Ans: D
Note that the inscribed circle of the $\Delta ABO$ is tangent to the $x$- and $y$- axis. Therefore, the distance between the centre and the $x$-axis and the distance between the centre and the $y$-axis is the same, i.e. the radius $r$. Hence, let $C(r,r)$ be the centre of the circle inscribed in $\Delta ABO$. Note that $D$, $E$ and $F$ are the points of contact of the inscribed circle and the $\Delta ABO$.
Note that the inscribed circle of the $\Delta ABO$ is tangent to the $x$- and $y$- axis. Therefore, the distance between the centre and the $x$-axis and the distance between the centre and the $y$-axis is the same, i.e. the radius $r$. Hence, let $C(r,r)$ be the centre of the circle inscribed in $\Delta ABO$. Note that $D$, $E$ and $F$ are the points of contact of the inscribed circle and the $\Delta ABO$.
Hence $OE=r$ and $OD=r$.
$\begin{array}{ll}
BF = BD & \text{(tangent properties)}\\
BF = OB – OD & \\
BF = 6 – r
\end{array}$
Similarly, $AF = 6-r$.
In $\Delta OAB$,
$\begin{array}{rcl}
AB^2 & = & OA^2 + OB^2 \\
AB & = & \sqrt{6^2 + 6^2} \\
AB & = & \sqrt{72}
\end{array}$
Hence, we have
$\begin{array}{rcl}
AB & = & AF + BF \\
\sqrt{72} & = & (6-r) + (6-r) \\
72 & = & (12-2r)^2 \\
4r^2 -48r +72 & = & 0 \\
r^2 – 12r + 18 & = & 0 \\
\end{array}$
Therefore, we have
$\begin{array}{rcl}
r & = & \dfrac{-(-12) \pm \sqrt{(-12)^2 – 4(1)(18)}}{2(1)} \\
& = & 6 \pm 3\sqrt{2} \\
\end{array}$
Since $r Hence, the coordinates of the in-centre of $\Delta ABO$ are $(6-3\sqrt{2},6-3\sqrt{2})$.
