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If two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8 x+2 y+8=0$ intersect in two distinct points, then
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Verified Answer
The correct answer is:
$2 < r < 8$
Centres and radii of the given circles are $C_1(1,3)$, $r_1=r$ and $C_2(4,-1), r_2=3$ respectively. Since, circles intersect in two distinct points, then
$$
\begin{aligned}
& \left|r_1-r_2\right| < C_1 C_2 < r_1+r_2 \\
& \Rightarrow \quad|r-3| < 5 < r+3 \\
&
\end{aligned}
$$
From last two relations, $r>2$
From first two relations,
$$
\begin{array}{ll}
& |r-3| < 5 \\
\Rightarrow & -5 < r-3 < 5 \\
\Rightarrow & -2 < r < 8
\end{array}
$$
$\therefore$ From Eq. (i) and Eq. (ii), we get
$$
2 < r < 8
$$
$$
\begin{aligned}
& \left|r_1-r_2\right| < C_1 C_2 < r_1+r_2 \\
& \Rightarrow \quad|r-3| < 5 < r+3 \\
&
\end{aligned}
$$
From last two relations, $r>2$
From first two relations,
$$
\begin{array}{ll}
& |r-3| < 5 \\
\Rightarrow & -5 < r-3 < 5 \\
\Rightarrow & -2 < r < 8
\end{array}
$$
$\therefore$ From Eq. (i) and Eq. (ii), we get
$$
2 < r < 8
$$
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