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  • Find the equation of tangent on Circle  at point  <img alt="(\sqrt{3},1)" src="ht

Find the equation of tangent on Circle x^2+y^2=4 at point  (\sqrt{3},1)

Option: 1

\sqrt{3}x+y=4


Option: 2

x+y=4


Option: 3

x+\sqrt{3}y=4


Option: 4

None of these


Answers (1)

best_answer

 

 

Equation of Tangent of Circle in Parametric Form -

Equation of Tangent of Circle in Parametric Form 

The equation of the tangent at the point to a circle is .

\text { If } \mathrm{S}=\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{a}^{2}=0 \text { is the circle, then the tangent at }\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is } \mathrm{T}_{1}=\mathrm{x} \mathrm{x}_{1}+\mathrm{yy}_{1}-\mathrm{a}^{2}=0

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x^2+y^2=4\\ \text{by above concept }\\ \text{The equation of the tangent at the point }(a \cos \theta, a \sin \theta)$ to a circle $x^{2}+y^{2}=a^{2}$ is $x \cos \theta+y \sin \theta=a\\ \text{so tangent at }(\sqrt{3},1) \text{here }\theta=\frac{\pi}{6}\\ x \cos {\frac{\pi}{6}}+y \sin\frac{\pi}{6} =2\\ \sqrt{3}x+y=4

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