Question

Consider the region $R=\{(x,y)\in R\times R:x\ge 0\text{and}{y}^2\le 4-x\}.$
Let $F$ be the family of all circles that are contained in $R$ and have centers on the $x-$axis. Let $C$ be the circle that has largest radius among the circles in $F.$ Let $(\alpha ,\beta )$ be a point where the circle $C$ meets the curve $y^2=4-x.$

The value of $\alpha$ is

Answer 1

Parabola : $y^2=4-x\dots \left(1\right)$
Circle : $(x-r{)}^2+y^2=r^2\dots (2)$
Solving $\left(1\right)$ and $\left(2\right),$ we get
$(x-r{)}^2+4-x=r^2$
$\Rightarrow x^2-(2r+1)x+4=0\cdots \left(3\right)$

At the common point of contact, equation $\left(3\right)$ has equal roots.
$\therefore (2r+1{)}^2-4^2=0$
$\Rightarrow (2r+1+4)(2r+1-4)=0$
$\Rightarrow r=\frac{3}{2}(\because r>0)$

Putting the value of $r$ in eqn. $\left(3\right),$ we get $x=\alpha =2$