Let $(h,k)$ be an external point from which tangents are drawn to the circle $x^2+y^2=a^2$. The center of this circle is $O=(0,0)$ and its radius is $a$. Let the angle between the tangents be $\alpha = 60^\circ$. The angle between the line segment from the external point to the center ($d = \sqrt{h^2+k^2}$) and one of the tangents is $\alpha/2 = 30^\circ$. In the right-angled triangle formed by the center, the external point, and the point of tangency, we have $\sin(\alpha/2) = \frac{\text{radius}}{\text{distance from center to external point}}$. So, $\sin(30^\circ) = \frac{a}{\sqrt{h^2+k^2}}$. Since $\sin(30^\circ) = 1/2$, we have $1/2 = \frac{a}{\sqrt{h^2+k^2}}$. Squaring both sides and cross-multiplying, we get $\sqrt{h^2+k^2} = 2a \Rightarrow h^2+k^2 = (2a)^2 = 4a^2$. Replacing $(h,k)$ with $(x,y)$, the locus is $x^2+y^2=4a^2$. This is the director circle for tangents subtending a $60^\circ$ angle.