The equation of the circle is, $(x - h)^2 + (y - k)^2$ = $r^2$ ....(i) Since the circle passes through point (2, 3) ∴ $(2 - h)^2 + (3 - k)^2$ = $r^2$ ⇒ 4 + $h^2$ – 4h + 9 + $k^2$ – 6k = $r^2$ ⇒ $h^2 + k^2$ – 4h – 6k + 13 = $r^2$ ....(ii) Also, the circle passes through point (–1, 1) ∴ $(-1 - h)^2 + (1 - k)^2$ = $r^2$ ⇒ 1 + $h^2$ + 2h + 1 + $k^2$ – 2k = $r^2$ ⇒ $h^2 + k^2$ + 2h – 2k + 2 = $r^2$ ....(iii) From (ii) and (iii), we have $h^2 + k^2$ – 4h – 6k + 13 = $h^2 + k^2$ + 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 = $\frac{7}{2}$ and k = $-\frac{5}{2}$ Putting these values of h and k in (ii), we get $\left(\frac{7}{2}\right)^2 + \left(-\frac{5}{2}\right)^2$ - $\frac{4 \times 7}{2} - 6 \times \frac{-5}{2}$ + 13 = $r^2$ ⇒ $\frac{49}{2} + \frac{25}{4} - 14 + 15 + 13$ = $r^2$ ⇒ $r^2$ = $\frac{65}{2}$ Thus required equation of circle is $\left(x - \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2$ = $\frac{65}{2}$ ⇒ $x^2 + \frac{49}{4} - 7x + y^2 + \frac{25}{4} + 5y$ = $\frac{65}{2}$ ⇒ $4x^2 + 49 - 28x + 4y^2$ + 25 + 20y = 130 ⇒ $4x^2 + 4y^2$ – 28x + 20y – 56 = 0 ⇒ 4($x^2 + y^2$ – 7x + 5y – 14) = 0 ⇒ $x^2 + y^2$ – 7x + 5y – 14 = 0.