Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes

Solution:

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

Since the circle passes through (0, 0) ,

(0 - h)² + (0 - k)² = r²

⇒ h² + k² = r²

The equation of the circle now becomes (x - h)² + (y - k)² = h² + k²

It is given that the circle makes intercepts a and b on the coordinate axes.

This means that the circle passes through points (a, 0) and (0, b)

Therefore,

(a - h)² + (0 - k)² = h² + k²

(0 - h)² + (b - k)² = h² + k²

From equation (1) , we obtain

⇒ a² + h² - 2ah + k² = h² + k²

⇒ a² - 2ah = 0

⇒ a (a - 2h) = 0

⇒ a = 0 or (a - 2h) = 0

However, a ≠ 0

Hence,

(a - 2h) = 0

⇒ h = a/2

From equation (2) , we obtain

h² + b² - 2bk + k² = h² + k²

⇒ b² - 2bk = 0

⇒ b (b - 2k) = 0

⇒ b = 0 or (b - 2k) = 0

However, b ≠ 0

Hence,

(b - 2k ) = 0

⇒ k = b/2

Thus, the equation of the required circle is

(x - a/2)² + (y - b/2)² = (a/2)² + (b/2)²

[(2x - a)/2]² + [(2y - b)/2]² = (a² + b²)/4

⇒ 4x² - 4ax + a² + 4y² - 4by + b² = a² + b²

⇒ 4x² + 4y² - 4ax - 4by = 0

⇒ 4 (x² + y² - ax - by) = 0

⇒ x² + y² - ax - by = 0

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NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 13

Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes

Summary:

The equation of the circle is x² + y² - ax - by = 0, passing through points (a, 0) and (0, b)