Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3)

Solution:

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

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x - h)² + y² = 25.

It is given that the circle passes through the point (2, 3).

Therefore,

⇒ (2 - h)² + 3² = 25

⇒ (2 - h)² = 25 - 9

⇒ (2 - h)² = 16

⇒ 2 - h = ± √16

⇒ 2 - h = ± 4

If, 2 - h = 4, then h = - 2

If 2 - h = - 4 , then h = 6

When h = - 2, the equation of the circle becomes

(x + 2)² + y² = 25

x² + 4x + 4 + y² = 25

x² + y² + 4x - 21 = 0

When h = 6, the equation of the circle becomes

(x - 6)² + y² = 25

x² - 12x + 36 + y² = 25

x² + y² - 12x + 11 = 0

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 12

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3)

Summary:

The equation of the circle are x² + y² + 4x - 21 = 0 and x² + y² - 12x + 11 = 0