Area of triangle in a circle. [Edexcel Specimen Papers Set 2, Paper 1H Q22]
The line $l$ is a tangent to the circle 
$^2 + ^2 = 40$ at the point $A$.  $A$ is the point $(2,6)$.
The line $l$ crosses the $x$-axis at the point $P$.
Work out the area of triangle $OAP$.
Any help is appreciated.Thank you.
I have attempted this - but I get the result $48$ (2sf) when the answer is $60$ units$^2$.
 A: You have the circle equation of
$$x^2 + y^2 = 40 \tag{1}\label{eq1A}$$
Implicitly differentiating it gives
$$\begin{equation}\begin{aligned}
2x + 2y\left(\frac{dy}{dx}\right) & = 0 \\
y\left(\frac{dy}{dx}\right) & = -x \\
\frac{dy}{dx} & = -\frac{x}{y}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, at $A(2,6)$, you have $\frac{dy}{dx} = -\frac{2}{6} = -\frac{1}{3}$. This is the slope of the tangent line. Using the equation form of $y = mx + b$ with the point $A$ then gives
$$\begin{equation}\begin{aligned}
6 & = -\frac{2}{3} + b \\
b & = \frac{20}{3}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, the line equation is
$$y = -\frac{x}{3} + \frac{20}{3} \tag{4}\label{eq4A}$$
To get the location where this line crosses the $x$-axis, set $y = 0$ in the equation above to get
$$0 = -\frac{x}{3} + \frac{20}{3} \implies x = 20 \tag{5}\label{eq5A}$$
As such, you have the point $P(20,0)$. Using the $\frac{bh}{2}$ formula for a triangle area gives that the requested area of $\triangle OAP$ is
$$\frac{20(6)}{2} = 60 \text{ units}^2\tag{6}\label{eq6A}$$
Since you haven't shown how you got $48$ instead, I'm not sure where your mistake might have been.
A: The slope of the radius from $O(0,0)$ to $A(2,6)$ is $\frac{6}{2}=3$.  The perpendicular slope of the tangent line at $(2,6)$ is the opposite reciprocal of the slope of $\overline{OA}$.  Therefore the slope of $\overline{AP}=-\frac{1}{3}$.
The equation of the tangent line is found by using the slope and the coordinate $A(2,6)$.
$$y-6=-\frac{1}{3}(x-2)$$
Plug in $y=0$ to find x-intercept (Point $P$):
\begin{align*}
0-6=-\frac{1}{3}(x-2)&\implies -6=-\frac{1}{3}(x-2)\\
&\implies (-3)(-6)=(x-2)\\
&\implies 18=x-2\\
&\implies x=20\\
\end{align*}
Therefore, point $P$ on the triangle is located at $P(20,0)$.
Take base of triangle to be $\overline{OP}=20~\mbox{units}$ and height of $\triangle{OAP}=6~\mbox{units}$.
Apply area formula: $A=\frac{1}{2}bh=\frac{1}{2}(20)(6)=60~\mbox{units}^2$
A: 
Let $|OA|=|OB|=r$, $|OC|=a$, $|AP|=t$, 
$|BP|=d$, $|AC|=h$.
Known values are: $h=6$, $a=2$, $r=\sqrt{40}$.
By Pythagorean theorem,
\begin{align}
\triangle AOP:\quad
r^2+t^2&=(r+d)^2
\tag{1}\label{1}
,\\
\triangle ACP:\quad
t^2&=
h^2+(d+r-a)^2
\tag{2}\label{2}
,\\
\end{align}
Substitution of $t^2$ from \eqref{2} into \eqref{1} gives
\begin{align}
r^2+h^2+(d+r-a)^2&=(r+d)^2
,
\end{align}
\begin{align}
d&=\frac{h^2+(r-a)^2}{2\,a}
=20-2\,\sqrt{10}
\tag{3}\label{3}
,
\end{align} 
Now the area $S$ of $\triangle AOP$
can be found as
\begin{align}
S&=\tfrac12\cdot|OP|\cdot|AC|
=\tfrac12\,(r+d)\,h
=60
.
\end{align}
