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A square is inscribed in the circle $x^2+y^2-2 x+4 y-93=0$ with its sides parallel to the co-ordinate axes. Then which among the following can be one of the vertices of the square?
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Verified Answer
The correct answer is:
$(8,5)$
Equation of given circle is
$$
\begin{gathered}
x^2+y^2-2 x+4 y-93=0 \\
\Rightarrow \quad(x-1)^2+(y+2)^2=98
\end{gathered}
$$
The vertex of the square inscribed in the circle whoses sides are parallel to the coordinate axes is
$$
\begin{aligned}
&\left(1 \pm \sqrt{98} \cos \frac{\pi}{4},-2 \pm \sqrt{98} \sin \frac{\pi}{4}\right) \\
&=(1 \pm 7,-2 \pm 7) \\
&=(8,5),(8,-9),(-6,5),(-6,-9)
\end{aligned}
$$
$$
\begin{gathered}
x^2+y^2-2 x+4 y-93=0 \\
\Rightarrow \quad(x-1)^2+(y+2)^2=98
\end{gathered}
$$
The vertex of the square inscribed in the circle whoses sides are parallel to the coordinate axes is
$$
\begin{aligned}
&\left(1 \pm \sqrt{98} \cos \frac{\pi}{4},-2 \pm \sqrt{98} \sin \frac{\pi}{4}\right) \\
&=(1 \pm 7,-2 \pm 7) \\
&=(8,5),(8,-9),(-6,5),(-6,-9)
\end{aligned}
$$
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