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If the number of common tangents to the pair of circles $x^2+y^2-2 x+4 y-4=0$ and $x^2+y^2+4 x-4 y+\alpha=0$ is 4 , then the least integral value of $\alpha$ is
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Verified Answer
The correct answer is:
5
We have,
$$
\begin{gathered}
x^2+y^2-2 x+4 y-4=0 \\
\text { and } x^2+y^2+4 x-4 y+\alpha=0 \\
c_1=(1,-2), r_1=\sqrt{1+4+4}=3 \\
c_2=(-2,2), r_2=\sqrt{4+4-\alpha}=\sqrt{8-\alpha}
\end{gathered}
$$
Circle have 4 common tangents
$$
\begin{aligned}
& \therefore c_1 c_2>r_1+r_2 \\
& \Rightarrow \quad \sqrt{(1+2)^2+(-2-2)^2}>3+\sqrt{8-\alpha} \\
& \Rightarrow \quad \sqrt{9+16}>3+\sqrt{8-\alpha} \\
& \Rightarrow \quad 2>\sqrt{8-\alpha} \\
& \Rightarrow \quad 8-\alpha < 4 \Rightarrow \alpha>4
\end{aligned}
$$
$\therefore$ Least integral value of $\alpha=5$
$$
\begin{gathered}
x^2+y^2-2 x+4 y-4=0 \\
\text { and } x^2+y^2+4 x-4 y+\alpha=0 \\
c_1=(1,-2), r_1=\sqrt{1+4+4}=3 \\
c_2=(-2,2), r_2=\sqrt{4+4-\alpha}=\sqrt{8-\alpha}
\end{gathered}
$$
Circle have 4 common tangents
$$
\begin{aligned}
& \therefore c_1 c_2>r_1+r_2 \\
& \Rightarrow \quad \sqrt{(1+2)^2+(-2-2)^2}>3+\sqrt{8-\alpha} \\
& \Rightarrow \quad \sqrt{9+16}>3+\sqrt{8-\alpha} \\
& \Rightarrow \quad 2>\sqrt{8-\alpha} \\
& \Rightarrow \quad 8-\alpha < 4 \Rightarrow \alpha>4
\end{aligned}
$$
$\therefore$ Least integral value of $\alpha=5$
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