The line L is a tangent to the circle *x ^{2 }+ y^{2 }= 50 *at the point A.

Line L crosses the x-axis at the point P.

Work out the area of triangle OAP. (O being the centre of the circle)

Guest Sep 13, 2017

#1**0 **

First, length of OA is \(\sqrt50\)as it's a radius of the circle. No need to simplify this,soon see why..

Now since line AP lies on the tangent line L ,then AP/OA = tan 90 deg

So AP/OA = 1

AP/\(\sqrt50 = 1\)

So AP is also \(\sqrt50 \)

Area of triangle is 1/2 base X height

= 1/2 X \(\sqrt50*\sqrt50\)

= 1/2 X 50

= 25

frasinscotland
Sep 13, 2017

#2**+2 **

Let's look at this a little closer......I don't know if "Point A" is some specific point or any random point on the circle.....but....if it's a random point.......look at this graph

Let " Point A " be on the on the circle = ( a , sqrt (r^2 - a^2))

The slope of the tangent line to this circle will be = -a / sqrt (r^2 - a^2)

So....the equation of a line tangent to the circle at this point is

y = [ -a / sqrt (r^2 - a^2)] ( x - a) + sqrt (r^2 - a^2)

To find "Point P" let y = 0 we have that

0 = [ -a / sqrt (r^2 - a^2)] ( x - a) + sqrt (r^2 - a^2)

[ a / sqrt (r^2 - a^2)] ( x - a) = sqrt (r^2 - a^2)

a (x - a) = r^2 - a^2

ax - a^2 = r^2 - a^2

ax = r^2

x = r^2 / a ... so "Point P" ( r^2/ a , 0 )

So....... the the base of triangle OAP will be r^2 / a and the height will be sqrt ( r^2 - a^2)

So....the area of this triangle will be (1/2) (r^2 / a) sqrt (r^2 - a^2) = (r^2) sqrt (r^2 - a^2) / [ 2a ]

So....suppose r = sqrt (50) and a = 4

Then the area of triangle OAP will be 50 sqrt (34) / 8 ≈ 36.44 sq units

But suppose that r =sqrt (50) and a = 2

Then the area of triangle OAP will be 50 sqrt ( 46) / 4 ≈ 84.78 sq units

CPhill
Sep 13, 2017