Answer:
The polar equation for the curve represented by the given cartesian equation is
![r=(10c)cos(\theta)](https://img.qammunity.org/2019/formulas/mathematics/college/y9kuq4hsk4cflsungawovkq8miy3jb0cf9.png)
Explanation:
We were given the following equation:
![x^2+y^2=10cx](https://img.qammunity.org/2019/formulas/mathematics/college/3zg0m03ooscosswqsxc7xsg9r2sktkjum0.png)
and the relations between cartesian and polar coordinates are given by
![x=rcos(\theta)](https://img.qammunity.org/2019/formulas/mathematics/college/a9rdrhgd48nydfaasmp70r8cevtsimssna.png)
![y=rsin(\theta)](https://img.qammunity.org/2019/formulas/mathematics/college/513ttdlb3ir8kwodb6ueruw7n2f8r6gmjd.png)
where r is a radius and θ an angle. Now we replace this relations in the original cartesian equation:
![(rcos(\theta))^2+(rsin(\theta))^2=(10c)rcos(\theta)\Leftrightarrow r^2cos^2(\theta)+r^2sin^2(\theta)=(10cr)cos(\theta)\Leftrightarrow r^2(cos^2(\theta)+sin^2(\theta))=(10cr)cos(\theta)](https://img.qammunity.org/2019/formulas/mathematics/college/lk7xmdjnt4fcythrb67jupk46fki008uh9.png)
and we use that
![(cos^2(\theta)+sin^2(\theta))=1](https://img.qammunity.org/2019/formulas/mathematics/college/2034mb37gt98fb4ldcw6ujf0si88r5229l.png)
to simplify, then
![r^2=(10c)rcos(\theta)\Leftrightarrow r=(10c)cos(\theta)](https://img.qammunity.org/2019/formulas/mathematics/college/tkr7wswwnpsdzzg0tkmpsruayac8zrp5zc.png)
wich is the polar equation for the curve.