Math, asked by WhatIsSpam, 2 months ago

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​

Answers

Answered by ItzMrSwaG
77

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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]

On solving eqn (3) and (4), we obtain h = 3 and k = 4.

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

(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 + y2² – 8y + 16 = 10

x² + y² – 6x – 8y + 15 = 0

  • https://brainly.in/question/34646638
Answered by Anonymous
0

Answer:

x^2 + y^2 – 6x – 8y + 15 = 0

Step-by-step explanation:

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

Since the circle passes through points (4, 1) and (6, 5),

(4 – h)^2 + (1 – k)^2 = r^2 …………………. (1)

(6 – h)^2 + (5 – k)^2 = r^2 …………………. (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 + h^2 + 1 – 2k + k^2 = 36 – 12h + h^2 + 25 – 10k + k^2

⇒ 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 = r^2

⇒ (1)^2 + (– 3)^2 = r^2

⇒ 1 + 9 = r^2

⇒ r^2 = 10

⇒ =√10

Thus, equation of the required circle is

(x – 3)^2 + (y – 4)^2 = (√10)^2

x^2 – 6x + 9 + y^2 – 8y + 16 = 10

x^2 + y^2 – 6x – 8y + 15 = 0

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