Question

NEED HELP ASAP, WILL GIVE BRAINLIEST

Answer 1

Given:

The equation of circle is

$(x+3)^2+y^2=9$

To find:

The polar form of given circle.

Solution:

We have,

$(x+3)^2+y^2=9$

$x^2+2(x)(3)+(3)^2+y^2=9$        $[\because (a+b)^2=a^2+2ab+b^2]$

$(x^2+y^2)+6x+9=9$

Subtracting 9 from both sides, we get

$(x^2+y^2)+6x=0$

We know that, $x^2+y^2=r^2$ and $x=r\cos \theta$.

$r^2+6(r\cos \theta)=0$

$r(r+6\cos \theta)=0$

$r=0$ and $r+6\cos \theta=0$

We know that, r is radius it cannot be 0. So,

$r+6\cos \theta=0$

$r=-6\cos \theta$

Therefore, the correct option is A.