The condition for circular motion is F=m\frac{v^2}{r} where \vec{v} is the tangential velocity.
And if the force is provided from gravitation force F=\frac{GmM}{r^2},
it impliy that v=\sqrt{\frac{GM}{r}}
Or m\frac{v^2}{r}=m\frac{(2\pi r/T)^2}{r}=m \frac{4\pi^2 r}{T^2}=\frac{GmM}{r^2}
so \frac{r^3}{T^2}=\frac{GM}{4\pi} which is a constant.

The trajectory of planet motion could be circular only if the above condition is satisfied, otherwise you might find an ellipse if the total energy is less than zero (closed orbit). 

Please check out the following related simulations at this web site:
[url=http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=766.0]planet motion for two stars[/url]
Circular motion: [url=http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=693.0]acceleration always perpendicular to velocity[/url]
[url=http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=629.0]How to determined the trajectory of a planetary motion?[/url]
[url=http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=547.0]motion of Moon, Earth relative to the Sun [/url]