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If the circle $x^2+y^2-6 x-8 y+\left(25-a^2\right)=0$ touches the axis of $x$, then a equals.
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Verified Answer
The correct answer is:
$\pm 4$
$\pm 4$
$$
\begin{aligned}
& x^2+y^2-6 x-8 y+\left(25-a^2\right)=0 \\
& \text { Radius }=4=\sqrt{9+16+\left(25-a^2\right)} \\
& \Rightarrow \mathrm{a}=\pm 4
\end{aligned}
$$
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