The equation of the circle is,
(x–h)2+(y–k)2 = r2 ....(i)
Since the circle passes through point (2, 3)
∴ (2–h)2+(3–k)2 = r2
⇒ 4 + h2 – 4h + 9 + k2 – 6k = r2
⇒ h2+k2 – 4h – 6k + 13 = r2 ....(ii)
Also, the circle passes through point (–1, 1)
∴ (–1–h)2+(1–k)2 = r2 ⇒ 1 + h2 + 2h + 1 + k2 – 2k = r2
⇒ h2+k2 + 2h – 2k + 2 = r2 ....(iii)
From (ii) and (iii), we have
h2+k2 – 4h – 6k + 13 = h2+k2 + 2h – 2k + 2
⇒ –6h – 4k = –11 ⇒ 6h + 4k = 11 .....(iv)
Since the centre (h, k) of the circle lies on the line x – 3y – 11 = 0.
∴ h – 3k – 11 = 0 ⇒ h – 3k = 11 ...(v)
Solving (iv) and (v), we get h =

7

2

and k = −

5

2

Putting these values of h and k in (ii), we get
(

7

2

)2+(−

5

2

)2 -

4×7

2

−6×

−5

2

+ 13 = r2
⇒

49

2

+

25

4

−14+15+13 = r2 ⇒ r2 =

65

2

Thus required equation of circle is (x−

7

2

)2+(y+

5

2

)2 =

65

2

⇒ x2+

49

4

−7x+y2+

25

4

+5y =

65

2

⇒ 4x2+49–28x+4y2 + 25 + 20y = 130 ⇒ 4x2+4y2 – 28x + 20y – 56 = 0
⇒ 4(x2+y2 – 7x + 5y – 14) = 0 ⇒ x2+y2 – 7x + 5y – 14 = 0.