SATHEE: Circle 3 Question 16

Circle 3 Question 16

16. Points $E$ and $F$ are given by

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

(b) $\frac{\sqrt{3}}{2}, \frac{1}{2}, (\sqrt{3}, 0)$

(d) $\frac{3}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, \frac{1}{2}$

Show Answer

Answer:

Correct Answer: 16. (a)

Solution:

Slope of line joining centre of circle to point $D$ is

$\tan θ = \frac{\frac{3}{2} - 1}{\frac{3 \sqrt{3}}{2} - \sqrt{3}} = \frac{1}{\sqrt{3}}$

It makes an angle $30^{\circ}$ with $X$ -axis.

$∴$ Points $E$ and $F$ will make angle $150^{\circ}$ and $- 90^{\circ}$ with $X$ -axis.

$∴ E$ and $F$ are given by

$\begin{aligned} \frac{x - \sqrt{3}}{\cos 150^{\circ}} & = \frac{y - 1}{\sin 150^{\circ}} = 1 \\ and & \frac{x - \sqrt{3}}{\cos (- 90^{\circ})} & = \frac{y - 1}{\sin (- 90^{\circ})} = 1 \\ ∴ & E & = \frac{\sqrt{3}}{2}, \frac{3}{2} and F = (\sqrt{3}, 0) \end{aligned}$

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Circle 3 Question 16

16. Points E and F are given by

Answer:

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16. Points $E$ and $F$ are given by