If the circle $x^2+y^2+2 \alpha x+c=0$ lies completely inside the circle $x^2+y^2+2 \beta x+c=0$, then which…

If the circle $x^2+y^2+2 \alpha x+c=0$ lies completely inside the circle $x^2+y^2+2 \beta x+c=0$, then which of the following holds?
  1. $\alpha \beta < 0$
  2. $c < 0$
  3. $c=0$
  4. $\alpha \beta>0$

Solution

The centre of the circle $x^2+y^2+2 \alpha x+c=0$ is $(-\alpha, 0)$. As the circle $x^2+y^2+2 \alpha x+c=0$ lies inside the circle $x^2+y^2+2 \beta x+c=0$, then $ \begin{aligned} & (-\alpha)^2+0^2+2 \beta(-\alpha)+c < 0 \\ & \alpha^2-2 \alpha \beta+c < 0 \\ & \alpha^2-2 \alpha \beta+\beta^2 < \beta^2-c \\ & (\alpha-\beta)^2 < \beta^2-c \end{aligned} $ Also, radius of circle $x^2+y^2+2 \alpha x+c=0$ less than radius of the circle $x^2+y^2+2 \beta x+c=0$ $ \begin{aligned} & \sqrt{\alpha^2-c} < \sqrt{\beta^2-c} \\ & \Rightarrow \alpha^2-c < \beta^2-c \Rightarrow \alpha < \beta \\ & \alpha^2-2 \alpha \beta < 0 \\ & \end{aligned} $ From Eqs. (i) and (ii), $ \begin{aligned} & c>0 \\ & \Rightarrow \quad-2 \alpha \beta < 0 \\ & \alpha \beta>0 \\ & \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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