Let $P$ be $\left(x_1, y_1\right)$
Now, equation of circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is
$\begin{aligned}
(x-r)^2+(y-r)^2 & =r^2 \\
\Rightarrow \quad x^2+y^2-2 x r-2 y r-r^2 & =0
\end{aligned}$
Now, polar of $P$ with respect to the circle
$\begin{aligned}
& x^2+y^2-2 x r-2 y r-r^2=0 \text { is } \\
& \quad x x_1+y y_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0 \\
& \Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r
\end{aligned}$
But it is given that polar of $P$ with respect to above circle is $x+2 y=4 r$
$\begin{aligned}
& \therefore \quad \frac{x_1-r}{1}=\frac{y_1-r}{2}=\frac{x_1+y_1-r}{4} \\
& \Rightarrow \quad x_1-r=\frac{x_1+y_1-r}{4} \text { and } x_1-r=\frac{y_1-r}{2} \\
& \Rightarrow \quad 4 x_1-4 r=x_1+y_1-r \text { and } 2 x_1-2 r=y_1-r \\
& \Rightarrow \quad 3 x_1-3 r=y_1 \text { and } 2 x_1-r=y_1 \\
& \therefore \quad 3 x_1-3 r=2 x_1-r \\
& \Rightarrow \quad \quad x_1=2 r \text { and } y_1=2 x_1-r=4 r-r=3 r \\
& \therefore \quad P \text { is }(2 r, 3 r)
\end{aligned}$