If a circle with the point $(-1,1)$ as its center touches the straight line $x+2 y+9=0$, then the coordinates of the points of contact is

The normal to the tangent will pass through the centre. Also, the intersection of the normal and tangent will give the point of contact. Hence, cosider the equation of tangent $$ y=-\frac{n}{2}-45 $$ Here, $\quad m=\frac{-1}{2}$ Therefore, slope of the normal is 2 . Since, slope of normal and tangent are mutually perpendicular. Therefore, equation of the normal at $(1,-1)$ is $$ \begin{aligned} & \frac{y-1}{x+1}=2 \text { or } y-1=2 x+2 \\ \Rightarrow \quad & 2 x-y=-3 \end{aligned} $$ Therefore, we have two equations $2 x-y=-3$ and $x+2 y+9=0$ Solving these two equations give us $(x, y)=(-3,-3)$ Hence, point of contact is $(-3,-3)$.