Answer:
focal length :
[tex]{ \tt{f = \frac{r}{2} }} \\ focal \: length = \frac{36}{2} = 18 \: cm[/tex]
From linear magnification:
[tex]{ \tt{ \frac{1}{m} + 1 = \frac{u}{f} }} \\ \frac{1}{9} + 1 = \frac{u}{18} \\ \\ u = \frac{18 \times 10}{9} \\ u = 20 \: cm[/tex]
The object must be at 20 cm
