Question

9. The line I is a tangent to the circle x? + y? = 40 at the point A A is the point (2, 6). The line I crosses the x-axis at the point P. Work out the area of triangle OAP do 날 L=69. The line I is a tangent to the circle x2 + y? = 40 at the point A A is the point (2, 6). The line I crosses the x-axis at the point P. Work out the area of triangle OAP

Answer 1

Explanation:

First, we need to find the coordinates of the center of the circle. Since the equation of the circle is x^2 + y^2 = 40, we have x^2 = 40 - y^2. Substituting this into the equation of the tangent line I, we get:

(40 - y^2) + y(2x) = 40

2xy - y^2 = 0

y(2x - y) = 0

Since the line I crosses the x-axis at point P, we have y = 0 at that point. Therefore, 2x - y = 0, and we can solve for x:

2x = y

2x = 6

x = 3

So the center of the circle is at the point (3, 0).

Next, we need to find the radius of the circle. Since A is on the circle, we have:

3^2 + 6^2 = r^2

r^2 = 45

r = sqrt(45) = 3sqrt(5)

Now we can find the coordinates of point O, which is the center of the circle:

O = (3, -3sqrt(5))

Finally, we can find the area of triangle OAP:

OA = sqrt((2-3)^2 + (6+3sqrt(5))^2) = sqrt(69)

AP = 2

Using the formula for the area of a triangle, we have:

Area(OAP) = 1/2 * OA * AP = 1/2 * sqrt(69) * 2 = sqrt(69)

Therefore, the area of triangle OAP is sqrt(69) square units.