How do you write the equation for a circle with center $(2a,a)$ and touching the y-axis?

Hint: In this question we will first find the radius of the circle and then use the standard equation of a circle and fit in the required variables and simplify it to get the required solution.

Formula used: Standard equation of a circle: ${(x - a)^2} + {(y - b)^2} = {r^2}$
Where $(a,b)$ is the center of the circle and $r$ is the radius of the circle

Complete step-by-step solution:
From the standard equation of the circle, we know that we need three things to get the equation of the circle.
We need the $x$ coordinate, $y$ coordinate and the radius of the circle which is represented as $r$.
In the question we have been given with the center of the circle, the center of the circle is $(2a,a)$
Which means the $x$ coordinate is $2a$and the $y$ coordinate is $a$.
Now it is given in the question that the circle touches the $y$ axis. This means that the radius of the circle is from the center till the $y$ axis.
Now the distance from the center to the $y$ axis is the $x$ coordinate of the center of the circle.
This means that the radius $r$ of the circle is $2a$ therefore $r = 2a$.
Now on substituting the values in the equation of the circle, we get:
$ \Rightarrow {(x - 2a)^2} + {(y - a)^2} = {(2a)^2}$
On simplifying the right-hand side of the equation, we get:
$ \Rightarrow {(x - 2a)^2} + {(y - a)^2} = 4{a^2}$

${(x - 2a)^2} + {(y - a)^2} = 4{a^2}$ is the required equation of the circle.

Note: It is to be noted that the value of the equation of the circle will change with the change in the value of $a$, which changes its center and radius.
The equation of a circle with its center at the origin is generally used, the equation is ${x^2} + {y^2} = {r^2}$
Where $r$ is radius of the circle