Question

Which of the following is the parametric form of the equation of the circle $x^2+y^2+px+py=0$

• A. $x=\frac{p}{2}(-1+\sqrt{2}\cos \theta )$; $y=\frac{p}{2}(-1+\sqrt{2}\sin \theta )$
• B. $x=\frac{p}{2}(1-\sqrt{2}\cos \theta )$; $y=\frac{p}{2}(-1-\sqrt{2}\sin \theta )$
• C. $x=\frac{p}{2}(-1-\sqrt{2}\cos \theta )$; $y=\frac{p}{2}(-1+\sqrt{2}\sin \theta )$
• D. None of these

Answer 1

The correct option is A $x=\frac{p}{2}(-1+\sqrt{2}\cos \theta )$; $y=\frac{p}{2}(-1+\sqrt{2}\sin \theta )$

The equation of the circle is $x^2+y^2+px+py=0$
This can also be written as $(x+\frac{p}{2}{)}^2+(y+\frac{p}{2}{)}^2=\frac{p^2}{2}$
If compared this equation in the form of $(x-h{)}^2+(y-k{)}^2=r^2,$ we get $h=-\frac{p}{2},k=-\frac{p}{2}$ and $r=\sqrt{\frac{p^2}{2}}$
Thus, in parametric form, $x=h+r\cos \theta$ and $y=r+k\sin \theta$ which gives
$x=-\frac{p}{2}+\sqrt{\frac{p^2}{2}},y=-\frac{p}{2}+\sqrt{\frac{p^2}{2}}$
i.e. $x=\frac{p}{2}(-1+\sqrt{2}\cos \theta )$; $y=\frac{p}{2}(-1+\sqrt{2}\sin \theta )$