Math/Precalculus

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?

Answer 1

x^2 + y^2 = 51^2 is the circle which gives the outer limits of the transmission

a straight line's equation from 71 miles north to 66 miles east will have an equation with x intercept 66 and y intercept 71, a line connecting those two points

(66,0) and (0,71) will have slope = -71/66

the equation of that line is y=-71x/66 + 71

find where that line intersects the circle. You'll get two points. calculate the distance between those 2 points

x^2 + (-71x/66 + 71)^2 = 51^2

x^2 + 5041x^2/4356 - 10082x/66 + 5041 = 2601

combine like terms and use the quadratic formula to solve for x

9397x^2/4356 - 10082x/66 + 2440 =0

x =10082/66)/(18794/4356) + or - (4356/18794)sqr(10082/66)^2 -4(9397/4356)(2440))

simplify and plug those values into the line's equation to solve for y

use the distance formula to find the length between those 2 points

if you don't make an arithmetic mistake somewhere, you're very good.

y=-66x/71 + 71 is a line connecting the intercepts with a midpoint (66/2, 71/2) =(33, 35.5)

33^2 + 35.5^2 = about 48.47^2< 51, so the circle should intersect the line twice, a little above and below that midpoint. You might try a rough sketch of the line and circle. It looks like only a short distance, that you're within 51 miles. a guess would be a one digit number, maybe less than 5.