The equation of a circle is of the type #x^2+y^2+2gx+2fy-c=0#. In this as the coefficients of #x^2# and #y^2# are equal and the term containing #xy# is not there, equation of a circle can be reduced to above form just by dividing by coefficient if #x# or #y#.
Now #x^2+y^2+2gx+2fy+c=0# can be rewritten as
#x^2+2gx+g^2+y^2+2fy+f^2+c=g^2+f^2# or
#(x+g)^2+(y+f)^2=g^2+f^2-c# or
#(x-(-g))^2+(y-(-f))^2=(sqrt(g^2+f^2-c))^2#
which is equation of a circle with center at #(-g,-f)# and radius #sqrt((g^2+f^2-c)#.
Hence, in the equation #x^2+5x+y^2-6y=3# or #x^2+y^2+5x-6y-3=0#, as #g=5/2#, #f=-3# and #c=-3#,
Center is #(-5/2,3)# and radius is #sqrt((-5/2)^2+3^2-(-3))#
= #sqrt(25/4+9+3)#
= #sqrt(73/4)=sqrt73/2#
Note - One can also make complete squares in the given equation directly to find center and radius, as follows
#x^2+5x+y^2-6y=3# or
#x^2+5x+25/4+y^2-6y+9=25/4+9+3# or
#(x+5/2)^2+(y-3)^2=73/4# or
#(x-(-5/2))^2+(y-3)^2=(sqrt73/2)^2#
and hence the result.
