Step 1:

Equation of the circle is $ x^2+y^2=x^2$ .

By symmetry, the arc length (circumference ) is 4x length of are in the arc length in the first quadrent

$L=4 \int \limits _0^a \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2 \normalsize dx}$

Step 2:

Now $x^2+y^2=a^2$

$\therefore \;2x+2ydy=0$

=>$\large\frac{dy}{dx}=\frac{-x}{y}$

Step 3:

$1+\bigg(\large\frac{dy}{dx}\bigg)$$=1+ \large\frac{x^2}{y^2}$

$\qquad=\large\frac{x^2+y^2}{y^2}$

$\qquad=\large\frac{a^2}{a^2-x^2}$

Step 4:

$L= 4 \int \limits_0^a \large\frac{a}{\sqrt {a^2-x^2}}\;$$dx$

$\quad= 4a \sin^{-1} \large\frac{x}{a}\bigg]_0^a $

$\quad=4a \times \large\frac{\pi}{2}$

$\quad=2 \pi a \;units $