Math/Precalculus/Circles

Question

A small radio transmitter broadcasts in a 51 mile radius. If you drive along a straight line from a city 71 miles north of the transmitter to a second city 66 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

miles=?

Answer 1

transmitting circle equation: x²+y²=51²=2601 (transmitter is at the centre)

A1(x=0, y=71), A2(x=66, y=0), thus the equation of the line connecting these two points is: y-71=-71/66(x-0)

Now, we have to find the contact points of the driving line and transmitting circle:

y=71-71/66(x)

x²+[71-71/66(x)]²=2601

x1=24.33, thus, y1=44.83

x2=46.4856, thus, y2= 21

These are the coordinates of the two points that we can receive the transmitter. The distance between these two points is:

d=√(y2-y1)²+(x2-x1)²= 32.53 miles

Answer 2

If you set up the x and y axis through the transmitter, then the equation of the circle of the transmitter's coverage becomes: x² + y² = 51² or x² + y² = 2601

The equation of the line of travel between the two cities becomes: y = -71/66x + 71

To find the intersection points, you can solve the system of equations using substitution:

x² + (-71/66x + 71)² = 2601

x² + 1.15725x² - 152.7576x + 5041 = 2601

2.15725x² - 152.7576x + 2440 = 0

Using the quadratic formula:

x = 24.34 and x = 46.47

To find the y-value, plug those in to y = -71/66x + 71 to get the intersection points: (24.34, 44.82) and (46.47, 21.01).

The distance between those points (using the distance formula) is √[(24.34 - 46.47)² + (44.82 - 21.01)²] = √(489.819 + 566.845) = 32.51

The total distance traveled (distance between the cities) is √(71² + 66²) = 96.94

So, you have radio coverage for about 32.5 miles or 32.51/96.94 = 34% of the drive