Question

Find the equation of the circle, whose centre is $(2,0)$ and touches the y axis.

Answer 1

The point $(2,0)$ is the centre of the required circle.

It is given that the circle touches the y-axis. Hence, the radius of the circle would be the shortest distance between its center and $x=0$.

$\therefore r={\left|\frac{x}{\sqrt{1}}\right|}_{x=2}=2$ (from distance between line and point)

The equation of the circle with centre $(h,k)$ and the radius $a$ is,

$(x-h{)}^2+(y-k{)}^2=a^2$

Hence, equation of circle is

$(x-2{)}^2+(y-0{)}^2=r^2$

$⟹x^2-4x+4+y^2=4$

$⟹x^2-4x+y^2=0$

This is the required equation.