Question 1038096
if you graph the two circles, you will see that the greatest distance will be from the center of the broadcast station, through the center of the transmitter, intersecting with the circumference of the transmitter.


that forms a straight line.


every other line emanating from the center of the transmitter will be at an angle from the line emanating from the center of the broadcast station, forming a triangle, whose largest leg will be shorter than the straight line.


the following graph shows you what i mean.


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AB is the straight line that starts from the center of the broadcast station and goes through the center of the transmitter and terminates on the circumference of the transmitter.


that's the line segment AB which is composed of line segment AD and DB


B is the center of the broadcast station.
D is the center of the transmitter.


the length of line AD is the radius of the circle of the transmitter, which is equal to sqrt(1600) which is equal to 40.


the length of line DB is equal to sqrt(30^2 + 40^2) = sqrt(2500) = 50.


the length of line AB is therefore 40 + 50 = 90.


that's the maximum range of the broadcast station, whose signal emanates from point B.


the equation of a circle is (x-a)^2 + (y-b)^2 = r^2


the center of the circle is (a,b).
the radius of the circle is sqrt(r^2) = r.


for the first transmitter, the center is at (0,0).
for the second transmitter, the center is at (-30,40)


the answer to the questions is:


the location of the television station is (0,0)
the location of the transmitter is (-30,40).
the maximum range of the first transmitter is 90 miles.


from the diagram, you can easily see that the length of AB is greater than the length of CB, which was drawn to show you that AB is the longest straight line that can be drawn from point B to the circumference of the transmitter.