Given that we need to find the equation of the circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes of co - ordinates.

Let us first find the points at which the line meets the axes.

The value of x is 0 on meeting the y - axis. So,

⇒ 3(0) + 4y = 12

⇒ 4y = 12

⇒ y = 3

The point is A(0,3)

The value of y is 0 on meeting the x - axis. So,

⇒ 3x + 4(0) = 12

⇒ 3x = 12

⇒ x = 4

The point is B(4,0)

We have the circle passing through the points O(0,0), A(0,3) and B(4,0).

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

⇒ x² + y² + 2ax + 2by + c = 0 ..... (1)

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

⇒ 0² + 0² + 2a(0) + 2b(0) + c = 0

⇒ c = 0 ..... (2)

Substituting A(0,3) in (1), we get,

⇒ 0² + 3² + 2a(0) + 2b(3) + c = 0

⇒ 9 + 6b + c = 0

⇒ 6b + c + 9 = 0 ..... (3)

Substituting B(4,0) in (1), we get,

⇒ 4² + 0² + 2a(4) + 2b(0) + c = 0

⇒ 16 + 8a + c = 0

⇒ 8a + c + 16 = 0 ..... (4)

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

⇒

Substituting these values in (1), we get

⇒

⇒ x² + y² - 4x - 3y = 0

∴The equation of the circle is x² + y² - 4x - 3y = 0.