NIMCET 2020 — Mathematics PYQ

Concept
Two curves $f(x,y)=0$ and $g(x,y)=0$ cut/touch at a point $(a,b)$ if $f(a,b)=g(a,b)=0$.
Calculation
Let f( x, y) = $\frac{x^2}{4}$+ $\frac{(y-1{)}^2}{9}$- 1= 0 and g( x, y) = $x^2$+ $y^2$- 4= 0 intersect at a point ( a, b) . 
$\therefore f(a,b)=g(a,b)=0$
$\Rightarrow \frac{{\mathrm{a}}^2}{4}+\frac{(\mathrm{b}-1{)}^2}{\mathrm{a}}-1={\mathrm{a}}^2+{\mathrm{b}}^2-4=0$ $\Rightarrow {\mathrm{a}}^2+\frac{4(\mathrm{b}-1{)}^2}{\mathrm{o}}-4={\mathrm{a}}^2+{\mathrm{b}}^2-4=0$ $\Rightarrow \frac{4(\mathrm{b}-1{)}^2}{9}={\mathrm{b}}^2$ $\Rightarrow 4($b$-1{)}^2=9{\mathrm{b}}^2$ $\Rightarrow (2\mathbf{b}-2{)}^2-(3\mathbf{b}{)}^2=0$
$\Rightarrow (2b-2+3b)(2b-2-3b)=0$
$\Rightarrow (5b-2)(-b-2)=0$
$\Rightarrow 5b-2=0$ OR-b-2=0
$$\Rightarrow \mathrm{b}=\frac{2}{5}\mathrm{OR}\mathrm{b}=-2.$$
Now,  using $a^2$+ $b^2$- 4= 0, we get: 
$${\mathrm{a}}^2+{\left(\frac{2}{5}\right)}^2-4=0{\mathrm{OR}\mathrm{a}}^2+(-2{)}^2-4=0$$
$\Rightarrow a^2=\frac{96}{25}$ OR $a^2$= 0

⇒ a = ±$\frac{4\sqrt{6}}{5}$ OR a = 0.
∴ The points of intersection are $\left(\frac{4\sqrt{6}}{5},\frac{2}{5}\right)$, $\left(-\frac{4\sqrt{6}}{5},\frac{2}{5}\right)$ and (0, -2).
The given ellipse and the circle intersect at three points (intersect at two and touch at one) as shown below: