Question

The circle $x^{2}+y^{2}=16$ tangent at the point $\theta=\frac{\pi}{3}$ is

$x+\sqrt{3}y=4$

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Answer

To determine the equation of the tangent to the circle $x^2 + y^2 = 16$ at a specific point given by $\theta = \frac{\pi}{3}$, follow the steps below:

  1. Identify the given circle and relevant parameters:

    • The equation of the circle is $x^2 + y^2 = 16$.
    • Recognize that this can be written as $x^2 + y^2 = a^2$ with $a = 4$ (since $4^2 = 16$).
  2. Calculate the coordinates of the tangency point:

    • Use parametric equations for the circle: $x = a \cos \theta$ and $y = a \sin \theta$.
      • Given $\theta = \frac{\pi}{3}$:
      • $x = 4 \cos \frac{\pi}{3} = 4 \cdot \frac{1}{2} = 2$
      • $y = 4 \sin \frac{\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$
  3. Determine the slope of the tangent line at the point:

    • The derivative of $y$ with respect to $x$ (dy/dx) for a circle is given by $\frac{dy}{d\theta} \div \frac{dx}{d\theta}$.

      • $x = 4 \cos \theta \Rightarrow \frac{dx}{d\theta} = -4 \sin \theta$
      • $y = 4 \sin \theta \Rightarrow \frac{dy}{d\theta} = 4 \cos \theta$
      • So, $\frac{dy}{dx} = \frac{4 \cos \theta}{-4 \sin \theta} = \frac{\cos \theta}{-\sin \theta} = -\cot \theta$
    • For $\theta = \frac{\pi}{3}$:

      • $\cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}$
      • Hence, $\frac{dy}{dx} = -\frac{1}{\sqrt{3}}$
  4. Form the equation of the tangent line:

    • Use the point-slope form of a line: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point $(2, 2\sqrt{3})$ and $m$ is the slope.
      • Here, $y - 2\sqrt{3} = -\frac{1}{\sqrt{3}} (x - 2)$
      • Simplify the equation: $$ y - 2\sqrt{3} = -\frac{1}{\sqrt{3}}x + \frac{2}{\sqrt{3}} $$ Multiply through by $\sqrt{3}$ to clear the fraction: $$ \sqrt{3} y - 6 = -x + 2 $$ Rearrange the terms: $$ x + \sqrt{3} y = 8 $$

Therefore, the correct equation of the tangent line is $x + \sqrt{3}y = 8$.


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