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$S \equiv x^2+y^2+2 x+3 y+1=0$ and
$S^{\prime} \equiv x^2+y^2+4 x+3 y+2=0$ are two circles.
The point $(-3,-2)$ lies
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$S^{\prime} \equiv x^2+y^2+4 x+3 y+2=0$ are two circles.
The point $(-3,-2)$ lies
Solution:
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Verified Answer
The correct answer is:
inside $S^{\prime}$ only
$S(-3,-2)=9+4-6-6+1=2>0$ $\therefore(-3,-2)$ outside of $S$ and $S(-3,-2)=9+4-12-6+2=-3 < 0$ $\therefore(-3,-2)$ inside of $S$.
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