JEE Main 2023 — Circle Question with Solution

Given region,

$$E=\left\{(x,y):3-x\le y\le \sqrt{9-x^2},0\le x\le 3\right\}$$

Since, point $(p,p+1)$ lie on line $y=x+1$

$\therefore$ Point of intersection of $y=x+1$ and $y=3-x$

i.e., $x+1=3-x$

$\Rightarrow$ $2x=2\Rightarrow x=1$

and $y=2$

and point of intersection of $y=x+1$ and

$$y=\sqrt{9-x^2}$$

i.e., $(x+1{)}^2=9-x^2$

$\begin{array}{lc} &\Rightarrow x^2+1+2 x=9-x^2 \\\\ &\Rightarrow 2 x^2+2 x-8=0 \\\\ &\Rightarrow x^2+x-4=0 \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{1+4(1)(4)}}{2} \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{17}}{2}\end{array}$

$\begin{array}{lll}\Rightarrow & x=\frac{-1+\sqrt{17}}{2},& \text{ (Since, }x\in [0,3])\\ & & \\ \therefore & p\in \left(1,\frac{-1+\sqrt{17}}{2}\right)& \end{array}$

$\Rightarrow a=1,b=\frac{-1+\sqrt{17}}{2}$

$\begin{array}{rl}& \therefore b^2+b-a\\ & \\ & =\frac{1+17-2\sqrt{17}}{4}+\frac{(-1+\sqrt{17})}{2}-1\\ & \\ & =\frac{18-2\sqrt{17}-2+2\sqrt{17}-4}{4}\\ & \\ & =\frac{12}{4}=3\end{array}$