Question

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

Answer 1

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

Since the centre of 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 … (1)

(0 – h)² + (b – k)² = h² + k² … (2)

From equation (1), we obtain

a² – 2ah + h² + k² = h² + k²

⇒ a² – 2ah = 0

⇒ a(a – 2h) = 0

⇒ a = 0 or (a – 2h) = 0

However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =.

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 =.

Thus, the equation of the required circle is