A square is inscribed in the circle $x^2+y^2-2x+8y-8=0$ whose diagonals are parallel to axes and a vertex in the first quadrant is A then OA is 
(1) 1                    (2) $\sqrt{2}$                     (3) $2\sqrt{2}$                      (4) 3

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+8\mathrm{y}-8=0 \\ \mathrm{Hence} \mathrm{centre} \mathrm{is} \left(1,-4\right) \\ \mathrm{Radius}=\sqrt{1^2+{\left(-4\right)}^2+\left(8\right)} \\ \mathrm{Hence} \mathrm{required} \mathrm{point} \mathrm{A} \mathrm{will} \mathrm{be} \mathrm{obtained} \mathrm{by} \mathrm{moving} 5 \mathrm{unit} \mathrm{up} \mathrm{from} \left(1,-4\right) \\ \mathrm{Hence} \\ \mathrm{A} \mathrm{is} \left(1,-4+5\right)=\left(1,1\right) \\ \mathrm{OA}=\sqrt{{\left(1-0\right)}^2+{\left(1-0\right)}^2}=\sqrt{2} \left[\mathrm{Answer}\right] \end{gathered}$$

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