Let the circle makes intercepts a with x-axis and b with y-axis.
∴ OA = a and OB = b
So the co-ordinates of A are (a, 0) and B are (0, b)
Now, the circle passes through three points O(0, 0), A(a, 0) and B(0, b).
Putting the co-ordinates of these three points in the general equation of circle
x2+y2 + 2gx + 2fy + c = 0 .....(i)
⇒ c = 0 (Since Circle passes through point O (0, 0))
Put (a, 0) in (i)
a2 + 2ga = 0 ⇒ a(a + 2g) = 0 ⇒ g = −

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2

a
Put (0, b) in (i)
b2 + 2fb = 0 ⇒ b(b + 2f) = 0 ⇒ f = −

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2

b
Putting these values of g, f and c in (i), we get
x2+y2+2×−

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2

x + 2×−

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2

by+0 = 0
⇒ x2+y2 – ax – by = 0
which is the required equation of circle