Answer
This is a cirlce with radius$\frac{5}{2}$ and center at $(\frac{5}{2},0)$.
Work Step by Step
$r=5cos\theta$
$r^{2}=5rcos\theta$
To convert polar equation to Cartesian coordinates, we use the equations $rcos \theta=x$ , $rsin \theta=y$ and $x^{2}+y^{2}=r^{2}$
Thus, $x^{2}+y^{2}=5x$
$(x^{2}-5x)+y^{2}=0$
Add $(\frac{5}{2})^{2}$ on both sides.
$(x^{2}-2 \cdot \frac{5}{2}x+(\frac{5}{2})^{2})+y^{2}=(\frac{5}{2})^{2}$
$(x-(\frac{5}{2}))^{2}+y^{2}=(\frac{5}{2})^{2}$
This is a cirlce with radius$\frac{5}{2}$ and center at $(\frac{5}{2},0)$.
