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.!
