Question

find the equation of circle passing through the points (1-4),and (5,-6) ha​

Answer 1

Step-by-step explanation:

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through 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 circle lies 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), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)2 + (1 – 4)2 = r2

⇒ (1)2 + (– 3)2 = r2

⇒ 1 + 9 = r2

⇒ r2 = 10

⇒

Thus, the equation of the required circle is

(x – 3)2 + (y – 4)2 =

x2 – 6x + 9 + y2 ­– 8y + 16 = 10

x2 + y2 – 6x – 8y + 15 = 0