Q.The parametric equations of the circle $x^2+y^2+mx+my=0$ are (A) $x=-\dfrac{m}{2}+\dfrac{m}{\sqrt2}\cos\theta,\ y=-\dfrac{m}{2}+\dfrac{m}{\sqrt2}\sin\theta$ (B) $x=-\dfrac{m}{2}+\dfrac{m}{\sqrt2}\cos\theta,\ y=+\dfrac{m}{2}+\dfrac{m}{\sqrt2}\sin\theta$ (C) $x=0,\ y=0$ (D) $x=m\cos\theta,\ y=m\sin\theta$

Complete the square on $x^2+y^2+mx+my=0$: $\left(x+\dfrac m2\right)^2-\dfrac{m^2}4+\left(y+\dfrac m2\right)^2-\dfrac{m^2}4=0$, i.e. $\left(x+\dfrac m2\right)^2+\left(y+\dfrac m2\right)^2=\dfrac{m^2}{2}$. So the centre is $\left(-\dfrac m2,-\dfrac m2\right)$ (both coordinates identical, since the equation is symmetric in $x$ and $y$) and the radius is $\sqrt{\dfrac{m^2}{2}}=\dfrac{m}{\sqrt2}$. Applying the shifted-centre parametric formulas $x=h+r\cos\theta,\ y=k+r\sin\theta$ with $h=k=-\dfrac m2$ and $r=\dfrac{m}{\sqrt2}$ gives identical-pattern equations for $x$ and $y$ (differing only in …

Complete the square (noting the equation is symmetric in $x,y$, so the centre has equal coordinates), then …

• Missing that the equation's symmetry forces the same shift and radius expression for both $x$ and $y$ (ruling out an option with a sign difference between the two) …