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(A) The lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $\frac{x}{a} - \frac{y}{b} = 1$.The vertices of the triangle are the intersection points of th

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$x^2 + y^2 - ax + by = 0$

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(A) The lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $\frac{x}{a} - \frac{y}{b} = 1$.The vertices of the triangle are the intersection points of these lines:$1$. Intersection of $x=0$ and $y=0$ is $O(0, 0)$.$2$. Intersection of $x=0$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $(0, -b)$.$3$. Intersection of $y=0$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $(a, 0)$.Since the triangle is a right-angled triangle with vertices at $(0,0)$,$(a,0)$,and $(0,-b)$,the hypotenuse is the line segment connecting $(a,0)$ and $(0,-b)$.The circumcircle of a right-angled triangle has the hypotenuse as its diameter.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting $(a, 0)$ and $(0, -b)$:$(x - a)(x - 0) + (y - 0)(y - (-b)) = 0$$x(x - a) + y(y + b) = 0$$x^2 - ax + y^2 + by = 0$$x^2 + y^2 - ax + by = 0$.

Find the equation of the circumcircle of the triangle formed by the lines $x = 0$,$y = 0$,and $\frac{x}{a} - \frac{y}{b} = 1$.

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