The equation of the circle is,
(x–h)2+(y–k)2 = r2 ....(i)
Since the circle passes through point (4, 1)
∴ (4–h)2+(1–k)2 = r2
⇒ 16 + h2 – 8h + 1 + k2 – 2k = r2
⇒ h2+k2 – 8h – 2k + 17 = r2 .... (ii)
Also, the circle passes through point (6, 5)
∴ (6–h)2+(5–k)2 = r2
⇒ 36 + h2 – 12h + 25 + k2 – 10k = r2
⇒ h2+k2 – 12h – 10k + 61 = r2 .... (iii)
From (ii) and (iii), we have
h2+k2 – 8h – 2k + 17 = h2+k2 – 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 = r2
∴ r2 = 10
Thus required equation of circle is (x–3)2+(y–4)2 = 10
⇒ x2 + 9 – 6x + y2 + 16 – 8y = 10 ⇒ x2+y2 – 6x – 8y + 15 = 0.