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# 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} = r^{2}

Since the circle passes through the 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 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 + 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

⇒ r = √10

Thus, the equation of the required circle is

(x - 3)^{2} + (y - 4)^{2} = (10)^{2}

x^{2} - 6x + 9 + y^{2} - 8 y + 16 = 10

x^{2} + y^{2} - 6x - 8 y + 15 = 0

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 x^{2} + y^{2} - 6x - 8 y + 15 = 0

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