Question:
A square is inscribed inthe circle
$x^{2}+y^{2}-6 x+8 y-103=0$ with its sides parallel to the corrdinate axes. Then the distance of the vertex of this square which is nearest to the origin is :-
Correct Option: , 4
Solution:
$\mathrm{R}=\sqrt{9+16+103}=8 \sqrt{2}$
$\mathrm{OA}=13$
$\mathrm{OB}=\sqrt{265}$
$\mathrm{OC}=\sqrt{137}$
$\mathrm{OD}=\sqrt{41}$
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