circle line of circle equation

Question

A small radio transmitter broadcasts in a 20 mile radius. If you drive along a straight line from a city 22 miles north of the transmitter to a second city 27 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

Answer 1

The answer to a previous problem like this was done using analytic geometry, so I will solve this using trig.

I will find the length of the road before reaching the signal and after losing the signal, and subtract this from the total distance.

We can create two triangles: the sides of the triangles include the city, the transmitter, and the point where the signal can first be/last be heard along the road. We will have the two distances and can determine, from the direction of the road, an angle, so we can solve. Note that, since the circumference of the circle would be perpendicular to the segment from the transmitter to the circle where it hits the road, the angle between the transmitter to the circle/road and the road will be obtuse.

I show lots of decimal places below. I will let you round as you like.°

Triangle 1: a = 22, b = 20, tan B = 27/22

Triangle 2: a = 27, b = 20, tan B = 22/27

(The tan of the angle in triangle 2 is the negative of the slope, as a is along the x-axis; it is the reciprocal of this, then, in triangle 2, as it is along the y-axis - the big triangle including the x and y axis and the road is, of course, a right triangle)

Triangle 1: B = tan^-1(27/22) = 50.8263420295558°

sin(A)/a = sin(B)/b, so A = 180° - sin^-1(22/20 * sin(50.8263420295558°)) = 180° - 58.512977018213 = 121.487022981787°

Then, C = 180° - A - B = 180° - 50.8263420295558° - 121.487022981787° = 7.68663498865766°

Then, as sin(C)/c = sin(B)/b, c = b sin(C)/sin(B) = 20 sin(7.68663498865766°)/sin(50.826420295558°) =

3.45069621757061

Triangle 2: B = tan^-1(22/27) = 39.1736579704442°

sin(A)/a = sin(B)/b, so A = 180° - sin^-1(27/20 * sin(39.1736579704442°)) = 180° - 58.512977018213 = 121.487022981787°. Note that this is the same angle as in triangle 1. You can see that this would be the case because we can construct a third triangle using the two points where the road hits circle and the transmitter, and this would be an isosceles triangle; the angles of interest are both supplements to the angle of interest.

Then, C = 180° - A - B = 180° - 39.1736579704442° - 121.487022981787° =

19.3393190477692°

Then, as sin(C)/c = sin(B)/b, c = b sin(C)/sin(B) = 20 sin(19.3393190477692°)/sin(39.1736579704442°) =

10.4852359009682.

Then, as the length of the drive is √ (22² + 27²) = 34.828149534536, we can pick up the signal for

34.828149534536 - 3.45069621757061 - 10.4852359009682 = 20.8922174159972

You can check these results by seeing that the x coordinates of the 2 points of intersection will be

27/√ (22² + 27²) * 3.45069621757061 and 27 - 27/√ (22² + 27²) * 10.4852359009682

and calculating y = sqrt(20^2 - x^2) - the reach of the transmitter signal - and y = 22 - 22/27 x, the point along the road, and seeing that the values are equal.