No, for the horse is in the 4th quadrant not the 1st quadrant, since
positive angles have their initial side on the right hand side of 
the x-axis and sweep around counter-clockwise from there.  Here is the
picture, where H marks the location of the particular horse:



We can ignore the outer circles as the horse we are talking about is on
the inner circle.  I suppose they told you about the other circles 
because there were other parts to the problem with horses on those circles,
right?  Anyway we'll erase the other circles and draw a radius vector
from the origin to the horse:



Since the point where the horse is located is 5/6 of the way around,
and ALL the way around is 360°, the angle swept out by the horse is
5/6 times 360° or 300°. 

Here is the 300°-arc through which the horse has traveled in going 5/6
of the way around the circle counter-clockwise from the right side
of the x-axis:



Polar coordinates are (r,q) where

r = DISTANCE TO ORIGIN,
q = ANGLE SWEPT OUT BY POINT TRAVELING FROM THE RIGHT SIDE OF THE
    x-AXIS, TAKEN POSITIVE IF SWEEPING IS DONE COUNTERCLOCKWISE,
    and NEGATIVE IF SWEEPING IS DONE CLOCKWISE.

Since the horse is 6 feet from the origin, r = 6, and the angle
swept out by the horse traveling counterclockwise from the right
side of the x-axis is 300°, then q = 300° and the polar coordinates 
of the point where the horse is located is

(r,q) = (6, 300°)




Edwin