Question

Find the eq of the circle passing through points (4, 1) and (6, 5) whose centre is on line 4x + y = 16.

urgent fast !!

Answer 1

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

✔️✔ 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)

✔️✔️centre (h, k) of the circle lies on line 4x + y = 16,

4h + k = 16 …... (3)

✔️✔️From (1) and (2),

(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)

solving (3) and (4),
h = 3
 k = 4.

✔️✔️On substituting the values of h and k in equation (1)
we get......

(4 – 3)2 + (1 – 4)2 = r2

⇒ (1)2 + (– 3)2 = r2

⇒ 1 + 9 = r2

⇒ r2 = 10

⇒r=√10

✔️✔️Thus, the equation of the required circle is:-

(x – 3)2 + (y – 4)2 =(√10)2

x2 – 6x + 9 + y2 ­– 8y + 16 = 10

⭕️x2 + y2 – 6x – 8y + 15 = 0⭕️

is the required equation

Answer 2

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Let say the equation 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]

Find the equation of the circle passing through the Points (4, 1)and (6, 5) and whose centre is on the line is 4x + y = 16