Find the equation of tangent in $1^{s t}$ quadrant to the circle $x^{2} + y^{2} = 16$ at $(2, 2 \sqrt{3})$

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Solution

The given circle is $x^{2} + y^{2} = 16$
The slope of tangent of circle is $2 x + 2 y \frac{d y}{d x} = 0 \frac{d y}{d x} = \frac{- x}{y} {\frac{d y}{d x} ∣ ∣ ∣}_{(2, 2 \sqrt{3})} = \frac{- 1}{\sqrt{3}}$

The equation of tangent is $y - 2 \sqrt{3} = \frac{- 1}{\sqrt{3}} (x - 2) \sqrt{3} y - 6 = - x + 2 x + \sqrt{3} y - 8 = 0$

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