The parametric equation of the circle ${{x}^{2}}+{{y}^{2}}+px+py=0$ is
(a) $x=\dfrac{-p}{2},y=\dfrac{-p}{2}$
(b) $x=\dfrac{p}{2}\cos \theta ,y=\dfrac{p}{2}\sin \theta $
(c) \[x=\dfrac{-p}{2}+\dfrac{p}{\sqrt{2}}\cos \theta ,y=\dfrac{-p}{2}+\dfrac{p}{\sqrt{2}}\sin \theta \]
(d) \[x=\dfrac{p}{\sqrt{2}}\cos \theta ,y=\dfrac{p}{\sqrt{2}}\sin \theta \]

Hint: Find the parametric equation of the circle for $x$ and $y$ respectively and the substitute the values of the centre and radius.

Complete step-by-step answer:
We know that the general equation of circle is of the form,
${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0....................\left( 1 \right)$
Now, we need to find the parametric equation of the circle.
For a circle with centre $\left( 0,0 \right)$ and radius $r$, the equation of circle is given as,
${{x}^{2}}+{{y}^{2}}={{r}^{2}}$
Similarly, for a circle with centre $\left( -g,-f \right)$, the equation is,
${{\left( x+g \right)}^{2}}+{{\left( y+f \right)}^{2}}+c=0............\left( 2 \right)$
Now let us expand the equation (2),
$\begin{align}
& {{x}^{2}}+2gx+{{g}^{2}}+{{y}^{2}}+2fy+{{f}^{2}}+c=0 \\
& {{x}^{2}}+2gx+{{y}^{2}}+2fy+{{g}^{2}}+{{f}^{2}}+c=0...........\left( 3 \right) \\
\end{align}$
Now, let us compare equation (1) with equation (3),
${{x}^{2}}={{g}^{2}}+{{f}^{2}}+c$
Thus, equation (3) can be modified as, ${{x}^{2}}+2gx+{{y}^{2}}+2fy+{{r}^{2}}=0$
Where the radius of the circle, $r=\sqrt{{{g}^{2}}+{{f}^{2}}+c}.........\left( 4 \right)$
The parametric equation of circle is given by,
Parametric equation of $X=-g+r\cos \theta $
$X=-g+\sqrt{{{g}^{2}}+{{f}^{2}}+c}\cos \theta $
(Where $\left( -g \right)$ is x-coordinate of centre of circle)
Parametric equation of $Y=-f+r\sin \theta $
$=-f+\sqrt{{{g}^{2}}+{{f}^{2}}+c}\sin \theta $
(where$\left( -f \right)$ is y-coordinate of the centre of the circle)
We got the parametric equation, now we need to find the values of $g,f,h.$
From the question, ${{x}^{2}}+{{y}^{2}}+px+py=0...........\left( 5 \right)$
Let us, compare equation (5) with equation (1)
In equation (1), the centre of the circle is $\left( -g,-f \right)$
In equation (5), the centre of the circle is $\left( \dfrac{-p}{2},\dfrac{-p}{2} \right)$,
Substitute these value on equation (1), i.e.
$\begin{align}
& {{x}^{2}}+{{y}^{2}}+2x\dfrac{p}{2}+2y\dfrac{p}{2}=0 \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}+px+py=0 \\
\end{align}$
Substitute $\left( \dfrac{-p}{2},\dfrac{-p}{2} \right)$ in equation (4)
$\therefore r=\sqrt{{{g}^{2}}+{{f}^{2}}+c}$
Take $c=0,$$\therefore r=\sqrt{{{\left( \dfrac{-p}{2} \right)}^{2}}+{{\left( \dfrac{-p}{2} \right)}^{2}}}=1$
i.e., value of
$\begin{align}
& g=\dfrac{-p}{2} \\
& f=\dfrac{-p}{2} \\
\end{align}$
$\therefore r=\sqrt{\dfrac{{{p}^{2}}}{4}+\dfrac{{{p}^{2}}}{4}}=\sqrt{\dfrac{2{{p}^{2}}}{4}}=\sqrt{\dfrac{{{p}^{2}}}{2}}=\dfrac{p}{\sqrt{2}}$
Substitute the values of $g=\dfrac{-p}{2},f=\dfrac{-p}{2},r=\dfrac{p}{\sqrt{2}}$ in the parametric equation of $X$ and $Y$.
$\therefore x=-g+r\cos \theta =-\dfrac{p}{2}+\dfrac{p}{\sqrt{2}}\cos \theta $
$y=-f+r\sin \theta =-\dfrac{p}{2}+\dfrac{p}{\sqrt{2}}\sin \theta $
Hence the correct option is (c).

Note: While solving this question, it's important how to use the parametric equation of the circle. Compare and get the values of the centre of circle and radius from the general equation of circle.