The parametric equations of the curve $x^2+y^2-a x-b y=0$ are

The parametric equations of the curve $x^2+y^2-a x-b y=0$ are
  1. $x=\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=-\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$
  2. $x=-\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$
  3. $x=-\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=-\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$
  4. $x=\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta, y=\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$

Solution

$\begin{aligned} & x^2+y^2-a x-b y=0 \\ & \Rightarrow\left(x-\frac{a}{2}\right)^2+\left(y-\frac{b}{2}\right)^2=\frac{a^2+b^2}{4} \end{aligned}$ which is circle having centre $\equiv(h, k) \equiv\left(\frac{a}{2}, \frac{b}{2}\right)$ $\text { and radius }=r=\sqrt{\frac{a^2+b^2}{4}}$ Its parametric equation is $\begin{aligned} & x=h+r \cos \theta=\frac{a}{2}+\sqrt{\frac{a^2+b^2}{4}} \cos \theta \\ & y=k+r \sin \theta=\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta \end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

Practice more Circle questions on Aicharya