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nekit [7.7K]
3 years ago
5

NEED HELP ASAP, WILL GIVE BRAINLIEST

Mathematics
1 answer:
Stella [2.4K]3 years ago
4 0

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.

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