Solve the 8th one

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Please find the below required solution.




$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{perpendicular} \mathrm{from} \mathrm{the} \mathrm{center} \mathrm{bisec} \mathrm{chord} \mathrm{so} \\ \mathrm{CM} = 2 \mathrm{cm} \\ \mathrm{and} \mathrm{AN}\hspace{0.17em}= 0.7 \mathrm{cm} \\ \mathrm{so} \mathrm{let} \mathrm{MO}\hspace{0.17em}= \mathrm{x} \mathrm{so} \mathrm{ON}\hspace{0.17em}= 3.9 - \mathrm{x} \\ \mathrm{now} \mathrm{radius} \mathrm{will} \mathrm{be} \mathrm{same} \\ \mathrm{so} \mathrm{using} \mathrm{pythagorus} \mathrm{theorem} \mathrm{in} \mathrm{triangle} \mathrm{CMO} \mathrm{and} \mathrm{ONM} \\ {\mathrm{CO}}^2\hspace{0.17em} = {\mathrm{OA}}^2 \\ {2}^2+{\mathrm{x}}^2={\left( 3.9-\mathrm{x}\right)}^2+0.49 \\ 4+{\mathrm{x}}^2= 15.21+{\mathrm{x}}^2-7.8\mathrm{x} +0.49 \left[\mathrm{using} {\left(\mathrm{x}-\mathrm{y}\right)}^2= {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}\right] \\ 7.8\mathrm{x} = 11.7 \\ \mathrm{x} = 1.5 \\ \mathrm{so} \mathrm{in} \mathrm{triangle} \mathrm{CMO} \\ \mathrm{CO}= \mathrm{radius} = \sqrt{{\mathrm{x}}^2+4}= \sqrt{2.25+4}= 2.5 \mathrm{cm} \end{gathered}$$

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