<s>[INST] 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$^2$.
 [/INST] 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: $A=\frac{1}{2}bh=\frac{1}{2}(20)(6)=60~\mbox{units}^2$
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