I will give #3# Methods to solve this problem.
Method I :-
Suppose that the reqd. eqn. of circle is # S : x^2+y^2+2gx+2fy+c=0#
Recall that the Centre #C# of #S# is #C(-g,-f)# which lies on
#y=5/2x+3 rArr -f=-5/2g+3...................(1)#.
Given that pt. #(2,5) in S rArr 4+25+4g+10f+c=0.....................(2)#.
#||ly, (6,7) in S rArr 36+49+12g+14f+c=0................(3)#
Solving #(1)-(3)# for #g,f and c# & using #S# will give us the reqd.
eqn. of circle # S : 9x^2+9y^2-44x-164y+647=0#.
Method II :-
Let us name the given pts. on #S# as #P(2,5) & Q(6,7)# and the
given line # l : y=5/2x+3......................(4)#.
Then, segment of line #PQ# is a chord of the Circle #S#. From
Geometry, we know that the centre #C# of #S# lies on the #bot#
bisector of every chord, in pparticular on that of #PQ#
We find the eqn. of the #bot# bisector #l'# of #PQ :-#
Slope of #PQ# is #(7-5)/(6-2)=2/4=1/2 rArr# slope of #l' bot PQ#
is#-2#, and, the mid-pt #M# of #PQ# lies on #l'#, i.e.,
#M((2+6)/2,(5+7)/2)=M(4,6) in l'#
#:. l' : y-6=-2(x-4)................................(5)#
Since, #l nn l' ={C}#, solving #(4) & (5)# gives #C#
Radius #r# of #S# is dist. #CP#
With the help of #C# and #r=CP#, we can derive the eqn. of #S#.
Method III :-
Chord #PQ# can be a Diameter of #S#. In the event, eqn. of a circle
having #P & Q# as diametrically opposite pts. is given by,
# S' : (x-2)(x-6)+(y-5)(y-7)=0#, or,
# S' : x^2+y^2-8x-12y+47=0......................(6)#
Eqn. of chord #PQ#, say #m#, having slope#=1/2# and #P(2,5)# on
it, is :#m: y-5=1/2(x-2)rArr2y-10=x-2rArrx-2y+8=0#.
We see that reqd. Circle #S# passes through #S' and m#, so, we
may assume that, # S : S'+lambdam=0, where, lambda in RR#, i.e.,
# S : x^2+y^2-8x-12y+47+lambda(x-2y+8)=0................(7)#
# S : x^2+y^2+(lambda-8)x-(2lambda+12)y+(8lambda+47)=0#.
Centre of this circle is #(-(lambda-8)/2, (2lambda+12)/2) in l#
# rArr lambda+6=-5/2((lambda-8)/2)+3#
# rArr 4lambda+24=-5lambda+40+12#
# rArr 9lambda=28#
# rArrlambda=28/9#
Using this with #(7)# we get #S#.
Enjoy Maths.!
