Given that we need to find the equation of the circle passing through the origin and cuts off intercepts a and b from x and y - axes.

Since the circle has intercept a from x - axis the circle must pass through (a,0) and (- a,0) as it already passes through the origin.

Since the circle has intercept b from x - axis, the circle must pass through (0,b) and (0, - b) as it already passes through the origin.

Let us assume the circle passing through the points O(0,0), A(a,0) and B(0,b).

We know that the standard form of the equation of the circle is given by:

⇒ x² + y² + 2fx + 2gy + c = 0 ..... (1)

Substituting O(0,0) in (1), we get,

⇒ 0² + 0² + 2f(0) + 2g(0) + c = 0

⇒ c = 0 ..... (2)

Substituting A(a,0) in (1), we get,

⇒ a² + 0² + 2f(a) + 2g(0) + c = 0

⇒ a² + 2fa + c = 0 ..... (3)

Substituting B(0,b) in (1), we get,

⇒ 0² + b² + 2f(0) + 2g(b) + c = 0

⇒ b² + 2gb + c = 0 ..... (4)

On solving (2), (3) and (4) we get,

⇒

Substituting these values in (1), we get

⇒

⇒ x² + y² - ax - by = 0

Similarly, we get the equation x² + y² + ax + by = 0 for the circle passing through the points (0,0), (- a,0), (0, - b).

∴The equations of the circles are x² + y²±ax±by = 0.