The equation of a circle that passes through the origin and cut off intercept -2 and 3 on the \(\mathrm{X}\)-axis and \(\mathrm{Y}\)-axis respectively is

The circle passes through the points \((0,0),(-2,0)\) and \((0,3)\).
We have \(x^2+y^2+2 g x+2 f y+c=0\)
\(
\therefore \mathrm{c}=0
\)
\(\ldots[\because\) It passes through \((0,0)]\)
\(
\therefore \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}=0
\)
\(
\therefore(-2)^2+2 \mathrm{~g}(-2)=0 \quad \Rightarrow 4-4 \mathrm{~g}=0 \Rightarrow \mathrm{g}=1
\)
\(
\text { Also }(3)^2+2 \mathrm{f}(3)=0 \quad \Rightarrow 6 \mathrm{f}=-9 \quad \Rightarrow \mathrm{f}=\frac{-3}{2}
\)
Thus centre \(\equiv\left(-1, \frac{3}{2}\right)\) and radius \(\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}\) Hence required equation of circle is
\(
x^2+y^2+2(1) x+2\left(\frac{-3}{2}\right) y+0=0 \text { i.e. } x^2+\) \( y^2+2 x-3y=0
\)