Circle Ques 56
- Points $E$ and $F$ are given by
(a) $\frac{\sqrt{3}}{2}, \frac{3}{2},(\sqrt{3}, 0)$
(b) $\frac{\sqrt{3}}{2}, \frac{1}{2},(\sqrt{3}, 0)$
(c) $\frac{\sqrt{3}}{2}, \frac{3}{2}, \frac{\sqrt{3}}{2}, \frac{1}{2}$
(d) $\frac{3}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, \frac{1}{2}$
Show Answer
Answer:
Correct Answer: 56.(a)
Solution:
- Slope of line joining centre of circle to point $D$ is
$$ \tan \theta=\frac{\frac{3}{2}-1}{\frac{3 \sqrt{3}}{2}-\sqrt{3}}=\frac{1}{\sqrt{3}} $$
It makes an angle $30^{\circ}$ with $X$-axis.
$\therefore$ Points $E$ and $F$ will make angle $150^{\circ}$ and $-90^{\circ}$ with $X$-axis.
$\therefore E$ and $F$ are given by
$$ \begin{array}{rlrl} & & \frac{x-\sqrt{3}}{\cos 150^{\circ}} & =\frac{y-1}{\sin 150^{\circ}}=1 \\ \text { and } & \frac{x-\sqrt{3}}{\cos \left(-90^{\circ}\right)} & =\frac{y-1}{\sin \left(-90^{\circ}\right)}=1 \\ \therefore & E & =\frac{\sqrt{3}}{2}, \frac{3}{2} \text { and } F=(\sqrt{3}, 0) \end{array} $$
BETA
