Question

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 (in units) is :

• A. $2$
• B. $3$
• C. $2\sqrt{2}$
• D. $3\sqrt{2}$

Answer 1

The correct option is C $2\sqrt{2}$

We know that, the equation of family of circles touching a line $L$ at $(x_1,y_1)$ is :
$(x-x_1{)}^2+(y-y_1{)}^2+\lambda L=0$
$\Rightarrow (x-1{)}^2+(y-1{)}^2+\lambda (x-y)=0$

The circle also passes through $(1,-3)$.
$\therefore 0+16+4\lambda =0\Rightarrow \lambda =-4$

Therefore, the equation of the circle is $x^2+y^2-6x+2y+2=0\Rightarrow r=\sqrt{(-3{)}^2+1^2-2}=2\sqrt{2}$

Alternate Solution:

Coordinates of the centre will be of form $(b,-1)$
Now equation of normal through $(1,1)$ is $x+y=2$
It will pass through the centre of the circle.
Hence $b=3$
$\therefore$ Radius $=\sqrt{2^2+2^2}=2\sqrt{2}$ units