Let us name the reqd. circle #S#, centre #C#, the given line #l : y=7/4x+4........(1)#, the given pts.#A(4,7)# & #B(2,5)#
#A and B in S rArr AB# is a chord of #S#.
From Geometry , we know that the #bot#bisector of every chord of a circle passes thro. the centre of the circle. Accordingly, #C# lies on #l'#, the #bot# bisector of chord #AB#.
Thus, #C in l'#, and by data, #C in l#, also. Therefore, to get #C#, we have to solve the eqns. of #l and l'#.
Eqn. of #l'# :-
The slope of chord #AB=(7-5)/(4-2)=2/2=1#
#l' bot l rArr# slope of #l'=-1#
The mid-pt.#M# of #AB# is #M((4+2)/2,(7+5)/2)=M(3,6)# & #M in l'#
#:. l' : y-6=-1(x-3)#, i.e., #y=6-x+3=9-x,............(2)#
Solving #(1) &(2)#, we get, # 7/4x+4=9-x#, i.e., #11/4x=5#, or, #x=20/11#
#:. y=9-x=9-20/11=79/11#
Therefore, #C=C(20/11,79/11)#
If #r# is the radius of #S#, then, with #C# as Centre and #B(2,5) in S#,
#r^2=CB^2=(2-20/11)^2+(79/11-5)^2=4/121+576/121=580/121#
Hence, # S : (x-20/11)^2+(y-79/11)^2=580/121#.
