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Question:
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$\begin{aligned}
& S \equiv x^2+y^2-2 x-4 y-4=0 \text { and } \\
& S^{\prime} \equiv x^2+y^2-4 x-2 y-16=0 \text { are two circles }
\end{aligned}$
the point $(-2,-1)$ lies
Options:
& S \equiv x^2+y^2-2 x-4 y-4=0 \text { and } \\
& S^{\prime} \equiv x^2+y^2-4 x-2 y-16=0 \text { are two circles }
\end{aligned}$
the point $(-2,-1)$ lies
Solution:
1817 Upvotes
Verified Answer
The correct answer is:
inside $S^{\prime}$ only
$\begin{gathered}
S(-2,-1)=(-2)^2+(-1)^2-2(-2)-4(-1)-4 \\
=4+1+4+4-4=9>0
\end{gathered}$
$\therefore(-2,-1)$ lies outside of $S$
$\begin{aligned}
S(-2,-1) & =(-2)^2+(-1)^2-4(-2)-2(-1)-16 \\
& =4+1+8+2-16=-1 < 0
\end{aligned}$
$\therefore(-2,-1)$ lies inside of $S^{\prime}$
Thus, $(-2,-1)$ lies inside $S^{\prime}$ only.
S(-2,-1)=(-2)^2+(-1)^2-2(-2)-4(-1)-4 \\
=4+1+4+4-4=9>0
\end{gathered}$
$\therefore(-2,-1)$ lies outside of $S$
$\begin{aligned}
S(-2,-1) & =(-2)^2+(-1)^2-4(-2)-2(-1)-16 \\
& =4+1+8+2-16=-1 < 0
\end{aligned}$
$\therefore(-2,-1)$ lies inside of $S^{\prime}$
Thus, $(-2,-1)$ lies inside $S^{\prime}$ only.
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