Find the equation of the circle which passes | vedclass

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(D) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$. Since the circle passes through the origin $(0,0)$,we have $c=0$. The circle cuts off a

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$x^2+y^2+2x-3y=0$

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(D) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$. Since the circle passes through the origin $(0,0)$,we have $c=0$. The circle cuts off an intercept of $-2$ on the $x$-axis,meaning it passes through $(-2,0)$. Substituting this into the equation: $(-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0$ $\Rightarrow 4 - 4g = 0$ $\Rightarrow g = 1$. The circle cuts off an intercept of $3$ on the $y$-axis,meaning it passes through $(0,3)$. Substituting this into the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0$ $\Rightarrow 9 + 6f = 0$ $\Rightarrow f = -\frac{3}{2}$. Substituting $g=1$,$f=-\frac{3}{2}$,and $c=0$ into the general equation: $x^2+y^2+2(1)x+2(-\frac{3}{2})y+0=0$. This simplifies to $x^2+y^2+2x-3y=0$.

Find the equation of the circle which passes through the origin and cuts off intercepts of $-2$ and $3$ on the $x$ and $y$ axes,respectively.

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