Let l be the fixed perimeter of a plane curve, between $x = x_{1}$ and $x = x_{2}$ then

$$l = \int\limits_{x_{1}}^{x_{2}} \sqrt{1 + y'^2}dx...........(1)$$

Let $A = \int\limits_{x_{1}}^{x_{2}} ydx............(2)$

Let F = y & $G = \sqrt{1 + y'^2}$

$$\therefore H = F + \lambda G \\ \therefore H = y + \lambda \sqrt{1 + y'^2} \\ \therefore \frac{\partial H}{\partial y} = 1, \frac{\partial H}{\partial y'} = \frac{\lambda y'}{\sqrt{1 + y'^2}}$$

H satisfies Euler's equation

$$\frac{\partial H}{\partial y} - \frac{d}{dx}\bigg(\frac{\partial H}{\partial y'} \bigg) = 0 \\ 1 - \frac{d}{dx} \frac{\lambda y'}{\sqrt{1 + y'^2}} = 0 \\ \frac{d}{dx} \frac{\lambda y'}{\sqrt{1 + y'^2}} = 1$$

Integrating above equation we get,

$$\frac{\lambda y'}{\sqrt{1 + y'^2}} = x - a$$

Squaring both sides we get,

$$(\lambda^2 - (x - a)^2)y'^2 = (x - a)^2 \\ y' = \pm \frac{x - a}{\sqrt{\lambda^2 - (x - a)^2}}$$

Integrating above equation we get,

$$y = \pm \sqrt{\lambda^2 - (x - a)^2} + b \\ \therefore (y - b) = \pm \sqrt{\lambda^2 - (x - a)^2}$$

Squaring both sides

$$(x - a)^2 + (y - b)^2 = \lambda^2$$ which is a circle.

Hence the curve is circle