In Cartesian coordinates, the generic circumference equation with center at point #p_0=(x_0,y_0)# and radius #r# is #(x-x_0)^2+(y-y_0)^2=r_0^2#
The pass equations are
#((x=r*cos(theta)),(y=r*sin(theta)))#
substituting we have
#(r*cos(theta)-x_0)^2+(r*sin(theta)-y_0)^2=r_0^2#.
Simplifying
#r^2-2r(x_0 cos(theta)+y_0 sin(theta))=r_0^2-x_0^2-y_0^2#
If now we call
#x_0=r_0 cos(theta_0)# and #y_0 = r_0 sin(theta_0)#
and substituting, after the identity
#cos(theta-theta_0)=sin(theta)sin(theta_0)+cos(theta)cos(theta_0)#
we get
#r^2+2r r_0 cos(theta-theta_0)-r_0^2+x_0^2+y_0^2#
Example.
Taking #(x-5)^2+(y-4)^2 = 1#
we have #x_0 = 5, y_0 = 4, r_0 = 1, theta_0 = arctan(4/5)#
so
#r^2+2 r cos(theta - theta_0) - 1+5^2+4^2 = 0#
or the other presentation
#(4 - r cos(theta))^2 + (5 - r sin(theta))^2=1#
