What are the coordinates of points E and F? | vedclass

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Question

(A) The slope of the line joining the center of the circle to point $D$ is $	an 	heta = \frac{3/2 - 1}{3\sqrt{3}/2 - \sqrt{3}} = \frac{1/2}{\sqrt{3}

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$\left( \frac{\sqrt{3}}{2}, \frac{3}{2} \right), (\sqrt{3}, 0)$

Vedclass · Answer

(A) The slope of the line joining the center of the circle to point $D$ is $	an 	heta = \frac{3/2 - 1}{3\sqrt{3}/2 - \sqrt{3}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.This line makes an angle of $30^{\circ}$ with the $x$-axis.Points $E$ and $F$ are at a distance of $1$ unit from the center $(\sqrt{3}, 1)$ at angles $30^{\circ} + 120^{\circ} = 150^{\circ}$ and $30^{\circ} - 120^{\circ} = -90^{\circ}$ respectively.For point $E$: $x = \sqrt{3} + 1 \cdot \cos(150^{\circ}) = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$ and $y = 1 + 1 \cdot \sin(150^{\circ}) = 1 + \frac{1}{2} = \frac{3}{2}$.For point $F$: $x = \sqrt{3} + 1 \cdot \cos(-90^{\circ}) = \sqrt{3} + 0 = \sqrt{3}$ and $y = 1 + 1 \cdot \sin(-90^{\circ}) = 1 - 1 = 0$.Thus,$E = \left( \frac{\sqrt{3}}{2}, \frac{3}{2} ight)$ and $F = (\sqrt{3}, 0)$.

What are the coordinates of points $E$ and $F$?

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