Here is a (way too) detailed walkthrough of this problem where I try to explain all the steps that you need to apply in order to solve such a problem using the Lagrange method. I will in this answer use the formalism laid out here where the constraints are separated out from the Lagrangian. This is different than what I mentioned in the comments above, but it's exactly the same (it's just a different packaging).
We always start by defining the Lagrangian and the constraints in the 'language' of the Lagrange method. The free Lagrangian for a point particle in 3D is given by
$$L = \frac{1}{2}m\left[\dot{x}(t)^2 + \dot{y}(t)^2 + \dot{z}(t)^2)\right]$$
where $\dot{x} \equiv \frac{dx}{dt}$. The constraint that the motion takes place in the plane, $x^2+y^2 = R^2$ with $z=0$, can be denoted by $H(x,y,z) = G(x,y,z) = 0$ where
$$G(x,y,z) = x^2 + y^2 - R^2$$
$$H(x,y,z) = z$$
The last constraint is trival in the sense that we can deal with it straight away since $z=0\to \dot{z} = 0$ so the $z$ dependence falls out and we can exclude $z(t)$ from the Lagrangian and effectively reducing the problem to a 2D problem with a single constraint $G(x,y) = x^2 +y^2 - R^2 = 0$.
After we have made the setup of the problem we need to calculate the Lagrange equations. The Lagrange's equations of the first kind are given by
$$\frac{dL}{dx} - \frac{d}{dt}\left[\frac{dL}{d\dot{x}}\right] - \lambda \frac{dG}{dx} = 0$$
$$\frac{dL}{dy} - \frac{d}{dt}\left[\frac{dL}{d\dot{y}}\right] - \lambda \frac{dG}{dy} = 0$$
To evaluate it we need the following terms
$$\frac{dL}{dx} = \frac{dL}{dy} = 0,~~~~\frac{dL}{d\dot{x}} = m\dot{x},~~~~\frac{dL}{d\dot{y}} = m\dot{y},~~~~\frac{dG}{dx} = 2x,~~~~\frac{dG}{dy} = 2y$$
Note that $\dot{x},\dot{y},x$ and $y$ are all threated as independent variables in the derivation so for example $\frac{d\dot{x}}{dx} = 0$ (which might seems strange at first).
Inserted into the Lagrange equations we get equations of motion for the particle
$$m\ddot{x} + 2\lambda x = 0$$
$$m\ddot{y} + 2\lambda y = 0$$
We can phrase this as a vector equation, by introducing $\vec{r} = (x,y)$, as
$$m\vec{a} = -2\lambda \vec{r}$$
where $\vec{a} = \frac{d^2\vec{r}}{dt^2} = (\ddot{x},\ddot{y})$ is the acceleration vector. From this form, and remembering Newton's second law, we see that the constraint $x^2 + y^2 = R^2$ that we imposed have given rise to a central-force $\vec{F} = -2\lambda\vec{r}$ on the particle. We still need to solve for $x,y$ and determine $\lambda$.
To solve for the evolution $(x(t),y(t))$ of the system we need to solve the Lagrange equations we derived above. The differential equations for $x$ and $y$ should be well known as those of a harmonic oscillator. The solutions reads
$$x(t) = A_x\sin(\omega t) + B_x\cos(\omega t)$$
$$y(t) = A_y\sin(\omega t) + B_y\cos(\omega t)$$
where $A_x,A_y,B_x,B_y$ are constants and $\omega^2 = \frac{2\lambda}{m}$. Note that $\lambda\geq 0$ is enforced (as we shall see below). To complete the solution we must enforce the constraint $x^2 + y^2 = R^2$ and put in the initial conditions.
Since the initial conditions are not given we simply pick our coordinate-system such that $x(0) = R$ giving $y(0) = 0$ (this is always possible). If you apply the constraint equation you will find $\dot{x}(0) = 0$ and find that the solution is characterized by only one free parameter $\dot{y}(0) \equiv v$. The constraints gives us $v = R\omega \implies 2\lambda = \frac{mv^2}{R^2}$. The full solution can now be written
$$x(t) = R\cos(\omega t)$$
$$y(t) = R\sin(\omega t)$$
which describes a particles moving around the circle with constant angular velocity. Since we have determined $\lambda$ in terms of $v$ we can also write the constraint-force in terms of $v$ as
$$\vec{F} = -\frac{mv^2}{R}\frac{\vec{r}}{R}$$
This is the well known expression for the centripetal force that makes a particles follow a circular path.
