Philip's method uses the tangent formula and Sue Liu of Madras College, St Andrew's sent a solution to this problem which depends on the use of Heron's Formula for the area of a triangle. Well done both of you.

FIRST METHOD

Using the tangents of the angles at the centre of the circle and the formula

\tan X = - \tan (A + B) = \frac{\tan A + \tan B}{\tan A \tan B - 1}

where

A + B + X = 180^{o}

\frac{x}{r} = \frac{\frac{a}{r} + \frac{b}{r}}{\frac{ab}{r^{2}} - 1} = \frac{r (a + b)}{ab - r^{2}} .

The area pf the triangle is

(a + b + x) r = (a + b) r + \frac{r^{3} (a + b)}{ab - r^{2}} = \frac{(a + b) abr}{ab - r^{2}}

as required.

SECOND METHOD
Here is Sue's method. The incircle divides the sides of the triangle into lengths

c' = a + b

b' = b + x

and

a' = x + a

as shown in the diagram. The semi-perimeter of the triangle is given by

s = x + a + b

and from Heron's formula the area of the triangle is

\begin{matrix} A & = & \sqrt{(s (s - a') (s - b') (s - c'))} \\ = & \sqrt{((a + b + x) abx)} \end{matrix}

Also, the triangle is divided into three smaller triangles and the total area is given by

$\begin{matrix} A & = & \frac{1}{2} (a + b) r + \frac{1}{2} (b + x) r + \frac{1}{2} (x + a) r \\ = & (a + b + x) r . \end{matrix}$

Equating the two answers

$\begin{matrix} \sqrt{abx (a + b + x)} & = & (a + b + x) r \\ abx (a + b + x) & = & (a + b + x)^{2} r^{2} \\ abx & = & (a + b + x) r^{2} \\ x & = & \frac{(a + b) r^{2}}{(ab - r^{2}) .} \end{matrix}$

Hence

$\begin{matrix} A & = & (a + b + x) r \\ = & \frac{r ((a + b) (ab - r^{2}) + (a + b) r^{2})}{ab - r^{2}} \\ = & \frac{abr (a + b)}{ab - r^{2} .} \end{matrix}$

An alternative method, not using Heron's formula, is based on finding x in terms of $a$ and $b$ using the tangents of the angles at the centre of the circle.