Question

The center of the circle $x=2+3\cos \theta ,y=3\sin \theta -1$ is

• A. $\left(3,3\right)$
• B. $\left(2,-1\right)$
• C. $\left(-2,1\right)$
• D. $\left(1,-2\right)$

Answer 1

The correct option is B

$\left(2,-1\right)$

Explanation for the correct option:

Finding the center of the circle:

Equation is given as $x=2+3\cos \theta ,y=3\sin \theta -1$

By rearranging them we get,

$\sin \theta =\frac{y+1}{3}$ and $\cos \theta =\frac{x-2}{3}$.

we know, ${\sin }^2\theta +{\cos }^2\theta =1$.

substituting the values $\sin \theta =\frac{y+1}{3}$ and $\cos \theta =\frac{x-2}{3}$.

$$\begin{gathered} \Rightarrow {\left(\frac{y+1}{3}\right)}^2+{\left(\frac{x-2}{3}\right)}^2=1 \\ \Rightarrow {\left(y+1\right)}^2+{\left(x-2\right)}^2=9 \end{gathered}$$

Cmparing this equation with equation of circle ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$. Here, $\left(a,b\right)$ is the center.

Hence, $\left(2,-1\right)$ is the center of the circle.

Therefore, the correct answer is option (B).