Question

How do you graph the point `D(-2,(13pi)/6)`?

Answer 1

See the small circle representing the point of zero radius, at D.
graph{(x+sqrt3)^2+(y+1)^2-.01=0 [-10, 10, -5, 5]}

Explanation:

In cartesian form, it is

`D( -2cos(13/6pi), -2sin(13/6pi))=D(-2cos(pi/6), -2sin(pi/6))`

`= D(-sqrt3, -1)`

Using Socratic utility, mark the point by a circle of small radius, with

center at D