# 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} = r^{2}

(2 - h)^{2} + (3 - k)^{2} = r^{2} ....(1)

(- 1 - h)^{2} + (1 - k)^{2} = r^{2} ....(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 + h^{2} + 9 - 6k + k ^{2} = 1 + 2h + h^{2} + 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} = r^{2}

⇒ [(4 - 7)/2]^{2} + [(6 + 5)/2]^{2} = r^{2}

⇒ (- 3/2) + (11/2) = r^{2}

⇒ 9/4 + 121/4 = r^{2}

⇒ 130/4 = r^{2}

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

4x^{2} - 28x + 49 + 4y^{2} + 20y + 25 = 130

4x^{2} + 4y^{2} - 28x + 20y - 56 = 0

4 (x^{2} + y^{2} - 7x + 5y - 14) = 0

x^{2} + y^{2} - 7x + 5y - 14 = 0

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 x^{2} + y^{2} - 7x + 5y - 14 = 0

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