The equation of the circle is, $(x – h)^2 + (y – k)^2$ = $r^2$ ....(i) Since the circle passes through point (4, 1) ∴ $(4 – h)^2 + (1 – k)^2$ = $r^2$ ⇒ 16 + $h^2$ – 8h + 1 + $k^2$ – 2k = $r^2$ ⇒ $h^2 + k^2$ – 8h – 2k + 17 = $r^2$ .... (ii) Also, the circle passes through point (6, 5) ∴ $(6 – h)^2 + (5 – k)^2$ = $r^2$ ⇒ 36 + $h^2$ – 12h + 25 + $k^2$ – 10k = $r^2$ ⇒ $h^2 + k^2$ – 12h – 10k + 61 = $r^2$ .... (iii) From (ii) and (iii), we have $h^2 + k^2$ – 8h – 2k + 17 = $h^2 + k^2$ – 12h – 10k + 61 ⇒ 4h + 8k = 44 ⇒ h + 2k = 11 ..... (iv) Since the centre (h, k) of the circle lies on the line 4x + y = 16 ∴ 4h + k = 16 ..... (v) Solving (iv) and (v), we get h = 3 and k = 4. Putting value of h and k in (ii), we get $(3)^2 + (4)^2$ – 8 × 3 – 2 × 4 + 17 = $r^2$ ∴ $r^2$ = 10 Thus required equation of circle is $(x – 3)^2 + (y – 4)^2$ = 10 ⇒ $x^2$ + 9 – 6x + $y^2$ + 16 – 8y = 10 ⇒ $x^2 + y^2$ – 6x – 8y + 15 = 0.