From the conversion formula, given by
![y=rsin\theta](https://img.qammunity.org/2023/formulas/mathematics/college/upbjftrym539nzqupeocp7rnbc02th6l7r.png)
we have that
![sin\theta=(y)/(r)](https://img.qammunity.org/2023/formulas/mathematics/college/mf57cfgoj6v7pujeqav0jc40fpvz7cugcf.png)
By substituting this result into the given polar equation, we have
![r=16*(y)/(r)](https://img.qammunity.org/2023/formulas/mathematics/college/5m1ptdbpymlf9wj2gbzszl0pyuyrgi4m7j.png)
So, by multiplying both sides by r, we get
![r^2=16y](https://img.qammunity.org/2023/formulas/mathematics/college/bma2xenlb2aadsxu8f8kmfb9nlyhaw36dl.png)
Now, from the conversion formula
![r^2=x^2+y^2](https://img.qammunity.org/2023/formulas/mathematics/college/wdjhwi4huh0mns5abr5vfni9ohfwt1o3nq.png)
we have that
![x^2+y^2=16y](https://img.qammunity.org/2023/formulas/mathematics/college/2ooe1fe1zik5yld21076knaffemgudnwwt.png)
By subtracting 16y to both sides, we obtain
![x^2+y^2-16y=0](https://img.qammunity.org/2023/formulas/mathematics/college/muvgeu52gh0sbf6rd38qyd1lpy1varc9a4.png)
By completing the square of the quadratic equation on y, we get
![x^2+(y-8)^2-64=0](https://img.qammunity.org/2023/formulas/mathematics/college/dzom2ttjvfp94ra5lt97apgrjwdj6zlght.png)
then, by adding 64 to both sides, we get
![x^2+(y-8)^2=64](https://img.qammunity.org/2023/formulas/mathematics/college/oit8wdiu914adky3y0jthh5ssrhj2f8w9j.png)
Therefore, the answer is the last option.