Let the equation of the circle
(x−h)2+(y−k)2=r2.............(A)
Since the circle passes through points (2,3) points (2,3) will satisfy the equation of a circle
Putting x=2,y=3 in (A)
(2−h)2+(3−k)2=r2(2)2+(h)2−2(2)(h)+(3)2+k2−2(3)(k)=r24+h2−4h+9+k2−6k=r2h2+k2−4h−6k+4+9=r2h2+k2−4h−6k+13=r2......(1)
Point (−1,1) will satisfy the equation of circle
Putting x=−1,y=−1 in (A)
(−1−h)2+(1−k)2=r2(−1)2+(1+h)2+(1−k)2=r2(1+h)2+(1−k)2=r2(1)2+h2+2h+1+k2−2k=r2h2+k2+2h−2k+1+1=r2h2+k2+2h−2k+2=r2......(2)
Since center (h,k) lie on the circle x−3y−11=0
Point (h,k) will satisfy the equation of line x−3y−11=0
So, h−3k−11=0
h−3k=11