Congratulations to Curt, Reigate College and to Andrei, Tudor Vianu National College, Bucharest, Romania for you solutions to this problem

When AB is a diameter angle $ACB$ is 90 degrees so we can use Pythagoras' Theorem. The area of the triangle is given by $\textstyle{1\over 2}ab = \textstyle {1\over 2}ch$ where $h$ is the length of the perpendicular from $C$ to $AB$. Then

$$(a+b)^2=a^2+b^2+2ab=c^2+2ch.$$

Thus $(a+b)^2$ is a maximum when $h$ is a maximum and equal to the radius of the circle $\textstyle {1\over 2}c$ . So the maximum value of $a+b$ is $c\sqrt 2$.

When $AB$ is not a diameter we have (using the Cosine
Rule):

$$\eqalign{ (a+b)^2 &= a^2 +b^2 + 2ab \cr &= c^2 +
2ab\cos \angle ACB +2ab \cr &= c^2 +2ab(1+\cos \angle ACB) \cr
&=c^2 +4ab \cos^2 \textstyle {1\over 2}\angle ACB.}.$$

As the area of triangle $ACB$ is given by $\Delta = \textstyle
{1\over 2} ab\sin \angle ACB$ we have

$$\eqalign { (a+b)^2 &= c^2 + {8\Delta \cos^2
\textstyle{1\over 2}\angle ACB \over \sin\angle ACB} \cr &= c^2
+ 4\Delta \cot \textstyle {1\over 2}\angle ACB }.$$

If we keep $A$ and $B$ fixed and vary $C$ then, as $c$ and
$\angle ACB$ are constant, and the area of the triangle $\Delta$ is
a maximum when $h$ is a maximum, it follows that $a + b$ is a
maximum when $h$ is a maximum, that is when $a=b$ and the altitude
of the triangle drawn from $C$ to $AB$ is a line of symmetry of the
triangle. In this case

$$a=b={c\over 2\sin \textstyle {1\over 2}\angle ACB}.$$

Conjecture: Let $Q$ be a variable cyclic quadrilateral in a
circle of radius $r$. Then the area and the perimeter of $Q$ will
be a maximum when $Q$ is a square; that is when each of the
diagonals of the quadrilateral is a diameter and each diagonal
bisects of the other diagonal at right angles.

Labelling $Q$ as $ABCD$, consider triangle $ABC$ with $AC$ fixed
and $B$ varying. Let $B'$ be the position of $B$ when, by the
previous result $AB + BC$ is a maximum, that is when $AB = BC$ Note
that this also gives the maximum area of triangle $ABC$. Similarly
by considering triangle $ADC$ with $AC$ fixed, we find $D'$ where
$AD+DC$ is a maximum and $AD=DC$. Now $B'D'$ is a diameter, keep
this fixed and consider triangles $B'CD'$ and $B'AD'$. The
positions of $C$ and $A$ that maximise the perimeter and area are
$C'$ and $A'$ where $A'C'$ is a diameter. Hence, for the perimeter
and area of $Q$ to be a maximum all the sides of the quadrilateral
must be equal making $Q$ a square. Hence the maximum perimeter is
$4r\sqrt 2$ and the maximum area is $2r^2$.