Question

The diameter of the circle whose centre lies on the line $x+y=2$ in the first quadrant and which touches both the lines $x=3$ and $y=2$ is

Answer 1

Let the $x-$ coordinate of centre of circle is $\alpha$ and radius $r.$
$\because$ centre lies on the $x+y=2$ and in $1^{st}$ quadrant.
$\Rightarrow \alpha +y=2\Rightarrow y=2-\alpha$
$\therefore$ centre of circle $=(\alpha ,2-\alpha )$
where $\alpha >0$ and $2-\alpha >0$
$\Rightarrow 0<\alpha <2$

$\because$ circle touches $x=3$ and $y=2$
$\Rightarrow |3-\alpha |=|2-(2-\alpha )|=$ r
$\Rightarrow |\alpha |=|\alpha -3|\Rightarrow {\alpha }^2=(\alpha -3{)}^2$
$\Rightarrow \alpha =\frac{3}{2}$
$\therefore$ $r=\alpha =\frac{3}{2}$
$\Rightarrow$ Diameter $\left(2r\right)=2\alpha =3$