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If the two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8 x+2 y+8=0$ intersect at two distinct points, then
Options:
Solution:
1250 Upvotes
Verified Answer
The correct answer is:
2 < r < 8
Equation of given circles
$$
\begin{aligned}
S_1:(x-1)^2+(y-3)^2 & =r^2 \\
\text { and } & S_2:(x-4)^2+(y+1)^2=3^2
\end{aligned}
$$
If circles $S_1$ and $S_2$ intersect at two distinct points, then
$$
\left|r_1-r_2\right| < C_1 C_2 < \left(r_1+r_2\right)
$$
$$
\begin{aligned}
& \text { So, }|r-3| < \sqrt{(4-1)^2+(-1-3)^2} < r+3 \\
& \Rightarrow(r-3) < 5 < r+3 \\
& \Rightarrow r+3>5 \text { and }|r-3| < 5 \\
& \Rightarrow r>2 \text { and } r-3 \in(-5,5) \\
& \Rightarrow r \in(-2,8) \\
& \Rightarrow 2 < r < 8
\end{aligned}
$$
Hence, option (a) is correct.
$$
\begin{aligned}
S_1:(x-1)^2+(y-3)^2 & =r^2 \\
\text { and } & S_2:(x-4)^2+(y+1)^2=3^2
\end{aligned}
$$
If circles $S_1$ and $S_2$ intersect at two distinct points, then
$$
\left|r_1-r_2\right| < C_1 C_2 < \left(r_1+r_2\right)
$$
$$
\begin{aligned}
& \text { So, }|r-3| < \sqrt{(4-1)^2+(-1-3)^2} < r+3 \\
& \Rightarrow(r-3) < 5 < r+3 \\
& \Rightarrow r+3>5 \text { and }|r-3| < 5 \\
& \Rightarrow r>2 \text { and } r-3 \in(-5,5) \\
& \Rightarrow r \in(-2,8) \\
& \Rightarrow 2 < r < 8
\end{aligned}
$$
Hence, option (a) is correct.
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