Question

The centre of the circle $r^2-2r(3cos\theta +4sin\theta )=9$ is

• A. $(5,ta{n}^{-1}\frac{4}{3})$
• B. $(5,ta{n}^{-1}\frac{3}{4})$
• C. $(\frac{5}{2},ta{n}^{-1}\frac{3}{4})$
• D. $(\frac{5}{2},ta{n}^{-1}\frac{4}{3})$

Answer 1

The correct option is A $(5,ta{n}^{-1}\frac{4}{3})$

Given, $r^2-2r(3\cos \theta +4\sin \theta )=9$
$\Rightarrow r^2-2\left(5\right)r\cos (\theta -\gamma )=9$ where, $\gamma ={\tan }^{-1}\left(\frac{4}{3}\right)$
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 circle $(5,{\tan }^{-1}\left(\frac{4}{3}\right))$
Ans: A