The line l is a tangent to the circle x^2 + y^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 the triangle OAP.

Firstly to work out the area of the triangle OAP we need to know the coordinates of the individual points O, A and P. O is the centre of the axis hence has coordinates (0,0). The coordinates of the point A have been given in the question as (2,6) and so we need to find the coordinates of the point P. As we are told that point P lies on the x-axis we know that the y-coordinate for P is 0. We also know the P lies on the tangent l and so to calculate the x-coordinate of P we need the equation of the line l. The general form of an equation for a straight line is y = mx + c, where m is the gradient and c is the y-intercept. To work out m directly we need two points on the line, as we are only given one which is A we must first find the gradient of the normal and then use that to find the gradient for the tangent. The two points which lie on the normal are O and A. So using m = y2-y1/x2-x1 we find that the gradient of the normal is 3. Using the knowldge that m of tangent x m of normal = -1 we can figure that the m for tangent is -1/3. Then using the coordinates for point A and the known m value we can calculate c as 20/3. Now as we have the equation for the line l which is y = -1/3(x) + 20/3 we can use that to find the x coordinate of P which is 20. Then we simply split OAP into two triangles and find the areas separately using 1/2 base x height and then add the two areas together to find the total area of the triangle OAP which is 60.