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# Find the equation of a circle with centre (2, 2) and passes through the point (4, 5)

**Solution:**

The centre of the circle is given as (h, k ) = (2, 2)

Since the circle passes through the point (4, 5),

the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

Therefore,

r = √(2 - 4)² + (2 - 5)²

= √(- 2)² + (- 3)²

= √4 + 9

= √13

Thus, the equation of the circle is

(x - h)^{2} + (y - k)^{2} = r^{2}

(x - 2)^{2} + (y - 2)^{2} = (√13)^{2}

x^{2} - 4x + 4 + y^{2} - 4 y + 4 = 13

x^{2} + y^{2} - 4x - 4 y - 5 = 0

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 14

## Find the equation of a circle with centre (2, 2) and passes through the point (4, 5)

**Summary:**

The equation of the circle is x^{2} + y^{2} - 4x - 4 y - 5 = 0

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