A circle of radius 5 units has diameter along the angle bisector of the lines $x+y=2$ and  $x-y=2$  & chord of contact from origin makes an angle of  ${45}^{\circ }$ with the positive direction of x-axis. If the equation of the circle is ${(x+\alpha )}^2+{(y+\beta )}^2=5^2$ then find  $|\alpha -\beta |$

Obviously angle bisectors are  $x=2$  and  $y=0$ . Now centre cannot lies on  $y=0$   because their chord of contact from origin will always be parallel to y-axis. So let the centre is  $(2,\alpha )$  then equation of circle will be

$\begin{array}{rr} & {(x-2)}^2+{(y-\alpha )}^2=5^2\\ & \Rightarrow x^2+y^2-4x-2\alpha y-21=0\end{array}$

Now chord of contact is

$\begin{array}{rr} & -4\frac{x}{2}-2\alpha \frac{y}{2}+{\alpha }^2-21=0\Rightarrow 2x+\alpha y-{\alpha }^2+21=0\\ & now -\frac{2}{\alpha }=1\Rightarrow \alpha =-2\end{array}$

So equation of circle is   ${(x-2)}^2+{(y+2)}^2=5^2$.