Answer:
[tex]area \: of \: the \: circle \: = \: \pi {r}^{2} \\ so \\ \: 10 = \pi {r}^{2} \\ {r}^{2} = \frac{10}{\pi} \\ r = \sqrt{ \frac{10}{\pi} } \\ r = \frac{ \sqrt{10} }{ \sqrt{\pi} } \\ r = \frac{ \sqrt{10} \times \sqrt{\pi} }{ \sqrt{\pi} \times \sqrt{\pi} } \\ r = \frac{ \sqrt{10\pi} }{\pi} \\ since \: d \: = 2r \\ so \\ d = 2( \frac{ \sqrt{10\pi} }{\pi} ) \\ \\ d = \frac{2 \sqrt{10\pi} }{\pi} \\ since \: circumferenc \: = 2r\pi \\ so \: circumference = 2( \frac{ \sqrt{10\pi} }{\pi} )\pi \\ = 2( \sqrt{10\pi} ) \\ = 2 \sqrt{10\pi} [/tex]
