Question -
Answer -
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passesthrough points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 … (1)
(6 – h)2 + (5 – k)2 = r2 … (2)
Since the centre (h, k) of the circlelies on line 4x + y = 16,
4h + k = 16 …(3)
From equations (1) and(2), we obtain
(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2
⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
⇒ 16 – 8h + 1 –2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k =44
⇒ h + 2k =11 … (4)
On solving equations (3) and (4), weobtain h = 3 and k = 4.
On substituting the values of h and k inequation (1), we obtain
(4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passesthrough points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 … (1)
(6 – h)2 + (5 – k)2 = r2 … (2)
Since the centre (h, k) of the circlelies on line 4x + y = 16,
4h + k = 16 …(3)
From equations (1) and(2), we obtain
(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2
⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
⇒ 16 – 8h + 1 –2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k =44
⇒ h + 2k =11 … (4)
On solving equations (3) and (4), weobtain h = 3 and k = 4.
On substituting the values of h and k inequation (1), we obtain
(4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
Thus, the equation of therequired circle is
(x – 3)2 + (y –4)2 = 
x2 – 6x + 9+ y2 – 8y + 16 = 10
x2 + y2 – 6x – 8y +15 = 0