Question

The centre of the circle $r^2-8r\cos (\theta -\frac{\pi }{6})=9$ is

• A. $(3,\frac{\pi }{6})$
• B. $(4,\frac{\pi }{6})$
• C. $(5,\frac{\pi }{6})$
• D. $(9,\frac{\pi }{6})$

Answer 1

Answer: (B)

$(4,\frac{\pi }{6})$

$r^2-8rcos(\theta -\frac{\pi }{6})=9$
$\Rightarrow r^2-2r\left(4\right)\cos (\theta -\frac{\pi }{6})+4^2=5^2$
On comparing with $r^2-2r{r}_0\cos (\theta -\gamma )+{r_0}^2=a^2$
where, $(r_0,\gamma )$ is center and $a$ is radius
Center of the circle is $(4,\frac{\pi }{6})$
Ans: B