Question

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

• A. (3, 3)
• B. (2, -1)
• C. (-2, 1)
• D. (-1, 2)

Answer 1

The correct option is B (2, -1)

x = 2 + 3 cos $\theta ,$ y = 3 sin $\theta$ - 1
$x^2+y^2=4+9co{s}^2\theta +12cos\theta +9si{n}^2\theta +1-6sin\theta$
$=14+12cos\theta -6sin\theta$
= 4 (2 + 3 cos $\theta$ ) - 2(3 sin $\theta$ - 1) + 4
$\Rightarrow x^2+y^2=4x-2y+4$
$\Rightarrow (x^2-4x+4)+(y^2+2y+1)=9$
$\Rightarrow (x-2{)}^2+(y+1{)}^2=9,\therefore$ centre is (2, -1).