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If \(y=\sqrt{3} x+k_1\) and \(y=\sqrt{3} x+k_2\) are two parallel tangents of a circle of radius 2 units, then \(\left|k_1-k_2\right|\) is equal to
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The correct answer is:
8
The distance between parallel tangents to a circle is equals to the diameter of the circle, so
\(\begin{aligned}
& \frac{\left|k_1-k_2\right|}{\sqrt{1+3}}=4 \\
\Rightarrow \quad & \left|k_1-k_2\right|=8
\end{aligned}\)
Hence, option (b) is correct.
\(\begin{aligned}
& \frac{\left|k_1-k_2\right|}{\sqrt{1+3}}=4 \\
\Rightarrow \quad & \left|k_1-k_2\right|=8
\end{aligned}\)
Hence, option (b) is correct.
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