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 =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