The line x = y touches a circle at the point (1, 1).If the circle also passes through the point $(1, -3)$,then its radius is :

 

 

 Equation of circle touching the line x = y at

 point $(1,1)$ is given by

$(x-1{)}^2+(y-1{)}^2+\lambda (x-y)=0$

Since it also passes through the point $(1,-3)$

$\therefore 0+(-3-1{)}^2+\lambda (1+3)=0\Rightarrow \lambda =-4$

$\therefore$ From $\text{ (i), }(x-1{)}^2+(y-1{)}^2-4(x-y)=0$

$\Rightarrow x^2+1-2x+1x^2+1-2y-4x-4y=0$

$\Rightarrow x^2+x^2-6x+2y+2=0$

$\therefore$Radius of the circle $=\sqrt{9+1-2}=2\sqrt{2}$ units