Question

The centre of the circle $r=a\cos \theta +b\sin \theta$ is

• A. ($\sqrt{a^2+b^2};ta{n}^{-1}\left(\frac{a}{b}\right))$
• B. ($\sqrt{a^2+b^2};ta{n}^{-1}\left(\frac{b}{a}\right))$
• C. $(\frac{\sqrt{a^2+b^2}}{2};ta{n}^{-1}(\frac{a}{b}\left)\right)$
• D. $(\frac{\sqrt{a^2+b^2}}{2};ta{n}^{-1}(\frac{b}{a}\left)\right)$

Answer 1

The correct option is D $(\frac{\sqrt{a^2+b^2}}{2};ta{n}^{-1}(\frac{b}{a}\left)\right)$

$r=a\cos \theta +b\sin \theta$
$r^2-r(acos\theta +bsin\theta )=0$
$\Rightarrow r^2-2\left(\frac{\sqrt{a^2+b^2}}{2}\right)r\cos (\theta -\gamma )=0$ where, $\gamma ={\tan }^{-1}\left(\frac{b}{a}\right)$
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 $(\frac{\sqrt{a^2+b^2}}{2},{\tan }^{-1}\left(\frac{b}{a}\right))$