Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0

Solution:

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

(2 - h)² + (3 - k)² = r² ....(1)

(- 1 - h)² + (1 - k)² = r² ....(2)

Since the centre (h, k) of the circle passes lies on the line x - 3y - 11 = 0,

h - 3k = 11 ....(3)

From equations (1) and (2) , we obtain

(2 - h)² + (3 - k)² = (- 1 - h)² + (1- k)²

⇒ 4 - 4h + h² + 9 - 6k + k ² = 1 + 2h + h² + 1 + k ² - 2k

⇒ 4 - 4h + 9 - 6k = 1 + 2h + 1 - 2k

⇒ 6h + 4k = 11 ....(4)

On solving equations (3) and (4), we obtain

h = 7/2 and k = - 5/2

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

⇒ (2 - 7/2)² + (3 + 5/2)² = r²

⇒ [(4 - 7)/2]² + [(6 + 5)/2]² = r²

⇒ (- 3/2) + (11/2) = r²

⇒ 9/4 + 121/4 = r²

⇒ 130/4 = r²

Thus, the equation of the required circle is

(x - 7/2)² + (y + 5/2)² = 130/4

[(2x - 7)/2]² + [(2y + 5)/2]² = 130/4

4x² - 28x + 49 + 4y² + 20y + 25 = 130

4x² + 4y² - 28x + 20y - 56 = 0

4 (x² + y² - 7x + 5y - 14) = 0

x² + y² - 7x + 5y - 14 = 0

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NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 11

Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0

Summary:

The equation of the required circle is x² + y² - 7x + 5y - 14 = 0