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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?
Options:
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Verified Answer
The correct answer is:
$\alpha \beta>0$
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),
$$
\begin{aligned}
& c>0 \\
& \Rightarrow \quad-2 \alpha \beta < 0 \\
& \alpha \beta>0 \\
&
\end{aligned}
$$
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),
$$
\begin{aligned}
& c>0 \\
& \Rightarrow \quad-2 \alpha \beta < 0 \\
& \alpha \beta>0 \\
&
\end{aligned}
$$
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