https://www.quora.com/What-is-the-equation-of-a-circle-which-passes-through-three-points-0-0-a-0-and-0-b
Because one point (0,0) is origin, one point (a,0) lies on x-axis and another (0,b) on y axis it is clear that is is a right angled triangle. so the circumcenter is the midpoint of the hypotenuse that is (\frac{a}{2},\frac{b}{2})
so the equation of the circle is
(x-\frac{a}{2})^2 +(y-\frac{b}{2})^2=(0-\frac{a}{2})^2 + (0-\frac{b}{2})^2 the RHS is square of distance of origin to to the centre
simplifying we get
x^2-ax+y^2-by=0
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