Q70 / 180 · COMEDK 2023 (Memory Based) · Mathematics

$\begin{aligned} & S \equiv x^2+y^2-2 x-4 y-4=0 \text { and } \\ & S^{\prime} \equiv x^2+y^2-4 x-2 y-16=0 \text { are two circles } \end{aligned}$ the point $(-2,-1)$ lies

  • Correct answer
    A
    inside $S^{\prime}$ only
  • B
    inside $S$ only
  • C
    inside $S$ and $S^{\prime}$
  • D
    outside $S$ and $S^{\prime}$

Solution

$\begin{gathered} S(-2,-1)=(-2)^2+(-1)^2-2(-2)-4(-1)-4 \\ =4+1+4+4-4=9>0 \end{gathered}$ $\therefore(-2,-1)$ 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.

Hence, the correct answer is option: 'A'.

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