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The point $(5,-7)$ lies outside the circle
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Verified Answer
The correct answer is:
$x^{2}+y^{2}-8 x=0$
Let $S_{1}: x^{2}+y^{2}-8 x=0$
$$
S_{2}: x^{2}+y^{2}-5 x+7 y=0
$$
$$
S_{3}: x^{2}+y^{2}-5 x+7 y-1=0
$$
$$
S_{4}: x^{2}+y^{2}-8 x+7 y-2=0
$$
Now, at $(5,-7)$
$$
\begin{aligned}
&S_{1}=25+49-40=34>0 \\
&S_{2}=25+49-25-49=0 \\
&S_{3}=25+49-25-49-1=-1 < 0 \\
&S_{4}=25+49-40-49-2=-17 < 0
\end{aligned}
$$
$\therefore(5,-7)$ lies outside the circle $S_{1}$.
$$
S_{2}: x^{2}+y^{2}-5 x+7 y=0
$$
$$
S_{3}: x^{2}+y^{2}-5 x+7 y-1=0
$$
$$
S_{4}: x^{2}+y^{2}-8 x+7 y-2=0
$$
Now, at $(5,-7)$
$$
\begin{aligned}
&S_{1}=25+49-40=34>0 \\
&S_{2}=25+49-25-49=0 \\
&S_{3}=25+49-25-49-1=-1 < 0 \\
&S_{4}=25+49-40-49-2=-17 < 0
\end{aligned}
$$
$\therefore(5,-7)$ lies outside the circle $S_{1}$.
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