1. The problem asks to find the radius of curvature $\rho$ at $\theta=0$ for the circle given in polar form: $$r = 2a \cos \theta.$$\n\n2. The general formula for radius of curvature in polar coordinates is: $$\rho = \frac{\left(r^2 + \left(\frac{dr}{d\theta}\right)^2\right)^{3/2}}{\left|r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r \frac{d^2r}{d\theta^2}\right|}.$$\n\n3. First, compute the derivatives of $r$ with respect to $\theta$:\n$$r = 2a \cos \theta,$$\n$$\frac{dr}{d\theta} = -2a \sin \theta,$$\n$$\frac{d^2r}{d\theta^2} = -2a \cos \theta.$$\n\n4. Evaluate these at $\theta=0$:\n$$r(0) = 2a \cos 0 = 2a,$$\n$$\frac{dr}{d\theta}(0) = -2a \sin 0 = 0,$$\n$$\frac{d^2r}{d\theta^2}(0) = -2a \cos 0 = -2a.$$\n\n5. Substitute into the formula for $\rho$:\n$$\rho = \frac{\left(r^2 + \left(\frac{dr}{d\theta}\right)^2\right)^{3/2}}{\left|r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r \frac{d^2r}{d\theta^2}\right|} = \frac{\left((2a)^2 + 0^2\right)^{3/2}}{\left|(2a)^2 + 2 \cdot 0^2 - (2a)(-2a)\right|}.$$\n\n6. Simplify numerator and denominator:\nNumerator: $$\left(4a^2\right)^{3/2} = (4a^2)^{1} \cdot \sqrt{4a^2} = 4a^2 \cdot 2a = 8a^3,$$\nDenominator: $$4a^2 + 0 + 4a^2 = 8a^2.$$\n\n7. Therefore,\n$$\rho = \frac{8a^3}{8a^2} = a.$$\n\n**Final answer:** The radius of curvature at $\theta=0$ is $\boxed{a}$.