MHTCET 11 May 2024 Shift 2 Paper

Concept:Convert the given circle equation into standard form $(x-h)^2+(y-k)^2=r^2$, then write its parametric equations using the center and radius.Explanation:Start with the equation:$$x^{2}+y^{2}-a x-b y=0$$Group the $x$ and $y$ terms and complete the squares:$$\left(x^{2}-a x+\frac{a^{2}}{4}\right)+\left(y^{2}-b y+\frac{b^{2}}{4}\right)=\frac{a^{2}}{4}+\frac{b^{2}}{4}$$This simplifies to:$$\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4}$$So the centre is $\left(\frac{a}{2},\frac{b}{2}\right)$ and the radius is:$$r=\frac{\sqrt{a^{2}+b^{2}}}{2}$$The parametric equations of a circle with centre $(h,k)$ and radius $r$ are:$$x=h+r\cos\theta,\qquad y=k+r\sin\theta$$Substitute the values:$$x=\frac{a}{2}+\frac{\sqrt{a^{2}+b^{2}}}{2}\cos\theta$$$$y=\frac{b}{2}+\frac{\sqrt{a^{2}+b^{2}}}{2}\sin\theta$$This matches option A.Answer:Option A: $x=\frac{a}{2}+\frac{\sqrt{a^{2}+b^{2}}}{2}\cos\theta,\quad y=\frac{b}{2}+\frac{\sqrt{a^{2}+b^{2}}}{2}\sin\theta$