Iso-parametric Dido problem in polar coordinates How to find maximum area arc between two points with fixed length, using polar coordinates?

I know that I can use Lagrange multipliers, but I cannot integrate the Euler-Lagrange equations. Is there not so messy solution when $ r(0)=r(\pi) $ and we want to maximize area above x axis?
 A: HINTS:
The answer is brief and includes essentials in polar coordinates derivation.
In rectangular coordinates the tangent to curve makes the angle $\phi$ made to x-axis. In polar coordinates $ (r,\theta)$ from origin O,$ \psi$ is angle the curve tangent makes to radius vector.
Primes are differentiation w.r.t. $\theta,$ the independent polar coordinate. Filling in the detailed steps is requested as inducing sufficient exercise in this polar coords case:
$$ \text{Lagrangian:  } \frac{r^2}{2}- \lambda \sqrt{r^2+r^{'2}}\tag1$$
Euler Lagrange Equation application when $\theta$ does not explicitly occur:
$$ \frac{r^2}{2}-\lambda \sqrt{r^2+r^{'2}}-r'( -\lambda \frac{r'}{\sqrt{r^2+r^{'2}}})= \text{const., set conveniently 
} =\frac{c^2}{2} \tag2$$
Using differential triangle trig relation as in the sketch
$$ \sin \psi=\frac{r}{\sqrt{r^2+r^{'2}}} \tag3$$
Simplifying (2)
$$ r \sin \psi= \frac{r^2-c^2}{2 \lambda} \tag4 $$
which is the differential equation of a Circle radius $\lambda $ and power $c^2$ in polar coordinates.
Geometric property verification
Two segment property of Circle at OAT. Drawing triangles ( not shown)  segment product is square of tangent or power:
$$ r (r-2 \lambda \sin \psi) = c^2 \tag 4 $$

When
$$ \psi=0, \text{there is tangential contact of radius vector at  T  i.e., when r=c.  } $$
We can place the Dido circle anywhere in x-y plane but we choose $ \psi =\pm \pi/2$ on x-axis. It is seen that $ \lambda $ is Dido circle /semi-circle radius and $c^2$ its power. Negative value for power is also admissible, when the origin is within the Dido Circle/semi- circle. Eccentric displacement $\epsilon= \sqrt{\lambda^2\pm c^2}.$
When
$$ r= \lambda+{\sqrt{r^2+r^{'2}}}=\lambda+{\sqrt{\lambda^2+c^2}}, \sin \psi =1, \psi = \pi/2 \quad  \text {at normal point Q    on substitution in 3) and simplification} $$
When
$$ r={\sqrt{r^2+r^{'2}}}- \lambda={\sqrt{\lambda^2+c^2}} -\lambda , \sin \psi =-1, \psi = -\pi/2\; \text {at normal point P on substitution in  3) and simplification.}$$
Integration
Eliminate $\psi$ between (3), (4) to obtain differential equation  $ r=f(\theta) $ form
$$ \frac{r'}{r}= \pm \frac{(r^2-c^2)} {\sqrt{2 r^2(2 \lambda^2+c^2) -r^4-c^4}} \tag 5 $$
which appears messier for analytic integration of / for the common Circle in polar coordinates compared to the Cartesian case.
