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

Solution:

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

Since the circle passes through the 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 the 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 + k ² = 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

⇒ 1² + (- 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 - 8 y + 16 = 10

x² + y² - 6x - 8 y + 15 = 0

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NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 10

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

Summary:

The equation of the required circle is x² + y² - 6x - 8 y + 15 = 0