JEE Main 2023 — Circle Question with Solution

Let the equation of circle touching both the coordinate axis be,

$(x-r{)}^2+(y-r{)}^2=r^2$ which passes through $\left(\alpha ,\beta \right)$

So, $(\alpha -r{)}^2+(\beta -r{)}^2=r^2 .......\left(1\right)$

Also equation of line $AB: x+y=r$

 

Let $Q\left(h,k\right)$,

Now using the formula of foot of perpendicular,

$\frac{h-\alpha }{1}=\frac{k-\beta }{1}=\frac{-(\alpha +\beta -r)}{2}$

$\Rightarrow (h,k)=\left(\alpha -\frac{\alpha }{2}-\frac{\beta }{2}+\frac{r}{2},\beta -\frac{\alpha }{2}-\frac{\beta }{2}+\frac{r}{2}\right)$

$\Rightarrow \left(h,k\right)=\left(\frac{\alpha -\beta +r}{2},\frac{-\alpha +\beta +r}{2}\right)$

Now using the distance formula we get,

${\left(PQ\right)}^2={\left(\frac{-\alpha -\beta +r}{2}\right)}^2+{\left(\frac{-\alpha -\beta +r}{2}\right)}^2=121$

$\Rightarrow (r-(\alpha +\beta ){)}^2=242$

$\Rightarrow r^2+(\alpha +\beta {)}^2-2r(\alpha +\beta )=242$

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta +r^2-2r\alpha -2r\beta =242$

$\Rightarrow 2\alpha \beta =242$ $\left\{\text{from equation} \left(1\right), (\alpha -r{)}^2+(\beta -r{)}^2=r^2\right\}$

$\Rightarrow \alpha \beta =121$