Question

I need some help on this question

Answer 1

`r=7\cos \theta \implies r^2=49\cos^2 \theta=7r \cos \theta \\ \\ \therefore x^2+y^2=7x \\ \\ x^2-7x+y^2=0 \\ \\ x^2-7x+(49)/(4)+y^2=(49)/(4) \\ \\ \left(x-(7)/(2) \right)^2+y^2=(49)/(4)`

So, the graph is a circle centered at
`(7/2, 0)` with a radius of 7/2. Thus, the graph is C.

The area is thus
`(7/2)^2 \pi=(49\pi)/(4)`