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If the circles $x^2+y^2-2 x-2(3+\sqrt{7}) y+8+6 \sqrt{7}=0$ and $x^2+y^2-8 x-6 y+k^2=0, k \in \mathbf{Z}$, have exactly two common tangents, then the number of possible values of $k$ is
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The correct answer is:
9
The centres of the given circles are $C_1(1,3+\sqrt{7})$ and $C_2(4,3)$ and corresponding radii are
$r_1=\sqrt{1^2+(3+\sqrt{7})^2-(8+6 \sqrt{7})}=3$
and $r_2=\sqrt{4^2+3^2-k^2}=\sqrt{25-k^2}$
Now, $C_1 C_2=\sqrt{(4-1)^2+(3-3-\sqrt{7})^2}=4$
Clearly, $C_1 C_2 < r_1+r_2$
$\begin{array}{lll}
\Rightarrow & 4 < 3+\sqrt{25-k^2} \Rightarrow 1 < \sqrt{25-k^2} & \\
\Rightarrow & 1 < 25-k^2 \Rightarrow k^2 < 24 \\
\therefore & k=0, \pm 1, \pm 2, \pm 3, \pm 4 \quad[\because k \in \mathbf{Z}]
\end{array}$
$r_1=\sqrt{1^2+(3+\sqrt{7})^2-(8+6 \sqrt{7})}=3$
and $r_2=\sqrt{4^2+3^2-k^2}=\sqrt{25-k^2}$
Now, $C_1 C_2=\sqrt{(4-1)^2+(3-3-\sqrt{7})^2}=4$
Clearly, $C_1 C_2 < r_1+r_2$
$\begin{array}{lll}
\Rightarrow & 4 < 3+\sqrt{25-k^2} \Rightarrow 1 < \sqrt{25-k^2} & \\
\Rightarrow & 1 < 25-k^2 \Rightarrow k^2 < 24 \\
\therefore & k=0, \pm 1, \pm 2, \pm 3, \pm 4 \quad[\because k \in \mathbf{Z}]
\end{array}$
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