The equation of a circle that passes through the origin and cut off intercept -2 and 3 on the…
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
$x^2+y^2-2 x+3 y=0$
$x^2+y^2+2 x+3 y=0$
$x^2+y^2+2 x-3 y=0$
$x^2+y^2-2 x-3 y=0$
Solution
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 $(0,0)]$
$\therefore \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}=0$
$\therefore(-2)^2+2 \mathrm{~g}(-2)=0 \quad \Rightarrow 4-4 \mathrm{~g}=0 \quad \Rightarrow \mathrm{g}=1$
$\text {Also }(3)^2+2 \mathrm{f}(3)=0 \quad \Rightarrow 6 \mathrm{f}=-9 \quad \Rightarrow \mathrm{f}=\frac{-3}{2}$
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
$x^2+y^2+2(1) x+2\left(\frac{-3}{2}\right) y+0=0 \text { i.e. } x^2+y^2+2 x-3 y=0$