Question 4866
A circle general equation is given by {{{(x-a)^2 + (y-b)^2 = r^2}}} where centre is (a,b) and radius = r.


So, for second quadrant, you want a centre having negative x and positive y, lets pick (-4,4) and a radius of 1...keep it simple :-)


So, we get {{{(x-(-4))^2 + (y-(4))^2 = 1^2}}} --> {{{(x+4)^2+(y-4)^2=1)}}}


As for semi-circle, choose your bounds for printing on the graph as x between -3 and -4, so it misses out the other half (-4 to -5).


Ok?


jon.