As all the points on a perpendicular bisector of a line segment joining two points #(x_1,y_1)# and #(x_2,y_2)#, where #x_1!=x_2# and #y_1!=y_2#, its equation would be
#(x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2#
or #x^2+y^2-2x x_1-2yy_1+x_1^2+y_1^2=x^2+y^2-2x x_2-2yy_2+x_2^2+y_2^2#
or #-2x x_1-2yy_1+x_1^2+y_1^2+2x x_2+2yy_2-x_2^2-y_2^2=0#
or #2x(x_2-x_1)+2y(y_2-y_1)+(x_1+x_2)(x_1-x_2)+(y_1+y_2)(y_1-y_2)=0#
(a) Hence equation of perpendicular bisector of line #k# joining #A(5,10)# and #B(-3,4)# is
#2x xx(-8)+2yxx(-6)+2xx8+14xx6=0# or #-16x-12y+100=0# i.e. #4x+3y=25#
and equation of perpendicular bisector of line #l# joining #A(5,10)# and #C(-1,-2)# is
#2x xx(-6)+2yxx(-12)+4xx6+8xx12=0# or #-12x-24y+120=0# i.e. #x+2y=10#
(b) We can find #knnl# by solving these simultaneous equations.
As #4xx(x+2y)-(4x+3y)=4xx10-25#, we have #5y=15# and #y=3# and putting this in say equation of #l#, #x+6=10# or #x=4#.
Hence #knnl=(4,3)# and this is the center of the circumcircle.
Radius is distance between center #(4,3)# and say #B(-3,4)#
#sqrt((-3-4)^2+(4-3)^2)=sqrt50# and hence
equation of circle is #(x-4)^2+(y-3)^2=50#
or #x^2+y^2-8x-6y-25=0#
