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 botbisector 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.
