Step 1: Test for symmetry about the x-axis by Replacing θ with -θ in the equation
The curve r=f(θ) is symmetrical about the x-axis if the equation r=f(θ) is unchanged by replacing θ with -θ
![r=5-5\cos (-\theta)=5-5\cos \theta\text{ (}\cos (-\theta)=\cos \theta)](https://img.qammunity.org/2023/formulas/mathematics/college/6btun34y9umgv1mup679w45oznj9gkpzvu.png)
Hence, the curve is symmetrical about the x-axis
Step 2 Test for symmetry about the y-axis by Replacing θ with -θ and r with -r in the equation
The curve r=f(θ) is symmetrical about the x-axis if the equation r=f(θ) is unchanged by replacing θ
with -θ and r with -r
![\begin{gathered} -r=5-5\cos (-\theta)=5-5\cos \theta \\ \text{ Therefore} \\ r=-5+5\cos \theta \\ \text{ Since } \\ -5+5\cos \theta\\e5-5\cos \theta \\ \text{ then there is no symmetry about the y-axis} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2fmr83a30cx62u5bffbgqhzff9df4o8qjl.png)
Step 2 Test for symmetry about the origin by replacing r with -r in the equation
The curve r=f(θ) is symmetrical about the origin if the equation r=f(θ) is unchanged by replacing r
with -r
![\begin{gathered} -r=5-5\cos \theta \\ \text{this implies that} \\ r=-5+5\cos \theta \\ \text{ Since} \\ -5+5\cos \theta\\e5-5\cos \theta \\ \text{ then there is no symmetry about the origin} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ernwooiaq94f27cv2ct2mzj1o7eupfzw6j.png)
Therefore, the curve is symmetrical only about the x-axis.
Hence, the right option is the first one.