Question

The centre of circle x = 3 + 2 cos∅, y=-1 + 2 sin∅ is

A) (1,-3)

B) (3,-1)

C) (3 , 1)

D) (1 ,3)​

Answer 1

Step-by-step explanation:

General equation of a circle is

${(x - a)}^{2} + {(y - b)}^{2} = {r }^{2}$

Where (a,b) is the co-ordinate of centre of the circle

Now,

$x = 3 + 2 \cos( \alpha ) \\ x - 3 = 2 \cos( \alpha )$

Now, squaring both side

${(x - 3)}^{2} = 4 {( \cos( \alpha )) }^{2} \: \: \: \: \: \: eqation \: (1)$

$y = - 1 + 2 \sin( \alpha ) \\ y + 1 = 2 \sin( \alpha )$

Squaring both side