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The parametric equations of the curve $x^2+y^2+a x+b y=0$ are
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Verified Answer
The correct answer is:
$x=-\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$
$$
\begin{aligned}
& x^2+y^2+\mathrm{a} x+\mathrm{b} y=0 \\
& \Rightarrow\left(x+\frac{\mathrm{a}}{2}\right)^2+\left(y+\frac{\mathrm{b}}{2}\right)^2=\frac{\mathrm{a}^2+\mathrm{b}^2}{4}
\end{aligned}
$$
Comparing with $(x-\mathrm{h})^2+(y-\mathrm{k})^2=\mathrm{r}^2$, we get
$$
\mathrm{h}=-\frac{\mathrm{a}}{2}, \mathrm{k}=-\frac{\mathrm{b}}{2}, \mathrm{r}=\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}}
$$
$\therefore \quad$ The parametric equations are
$$
x=-\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta
$$
\begin{aligned}
& x^2+y^2+\mathrm{a} x+\mathrm{b} y=0 \\
& \Rightarrow\left(x+\frac{\mathrm{a}}{2}\right)^2+\left(y+\frac{\mathrm{b}}{2}\right)^2=\frac{\mathrm{a}^2+\mathrm{b}^2}{4}
\end{aligned}
$$
Comparing with $(x-\mathrm{h})^2+(y-\mathrm{k})^2=\mathrm{r}^2$, we get
$$
\mathrm{h}=-\frac{\mathrm{a}}{2}, \mathrm{k}=-\frac{\mathrm{b}}{2}, \mathrm{r}=\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}}
$$
$\therefore \quad$ The parametric equations are
$$
x=-\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta
$$
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