Find the equation of the normal to the circle $x^2+y^2=2x$, which is parallel to the line $x+2y=3$.

 

Given circle is $x^2+y^2-2x=0$

Centre of given circle is $(1,0)$

Since, normal is parallel to $x+2y=3$

let the equation of normal is $x+2y=\lambda$

Since, normal passes through the center of the circle $(1,0)$ then

$\Rightarrow 1+0=\lambda \Rightarrow \lambda =1$

Then, the equation of the normal is $x+2y-1=0$.