If the line $y = \sqrt[]{3} x + k$ touches the circle $x^{2} + y^{2} = 16$ , then find the value of $k$ .

[Hint: Equate perpendicular distance from the center of the circle to its radius.]

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Alok Kumar Singh | Contributor-Level 10

7 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n c i r c l e i s x^{2} + y^{2} = 16 \\ P e r p e n d i c u l a r f r o m t h e o r i g i n t o t h e g i v e n l i n e y = \sqrt[]{3} x + k i s e q u a l t o t h e r a d i u s . \\ ∴ 4 = | \frac{0 - 0 - k}{\sqrt[]{{(1)}^{2} + {(\sqrt[]{3})}^{2}}} | = | \frac{- k}{4} | \\ \Rightarrow 4 = \pm \frac{k}{2} \Rightarrow k = \pm 8 \\ H e n c e, t h e r e q u i r e d v a l u e s o f k a r e \pm 8 . \end{array}$

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