Circle equation

Question

A small radio transmitter broadcasts in a 25 mile radius. If you drive along a straight line from a city 32 miles north of the transmitter to a second city 29 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

Answer 1

We can place the radio transmitter at (0,0) and radius 25 so the
equation of the circle is: x² + y² = 25²  ...  x² + y² = 625

 

We drove solely from point (0, 32) to point (29,0) ..

The slope of this line is (-32/29) and y-intersection at (0,32).

The equation of the line is  y = (-32/29)x + 32

The distance driven is √(29² +(-32)²) ≅ 43.18565 miles
along the line y = (-32/29)x + 32
 

We need to check for intersections between the equations:
x² + y² = 625
y = (-32/29)x + 32 
 
If there are any, we need to find the distance between the
intersection points to determine the number of miles of the
total distance of 43.2 miles was within the circle.
 

Answer 2

1. Determine the equation of the path driven

2. Determine the equation of the circle

3. Solve for the points of intersection.

4. Calculate the distance between the two points

Answer 3

We can put transmitter AND the center of the circle at (0,0)

The radius is 25.

 

The equation of the circle is x^2 + y^2 = 625

 

The distance driven is 32 + 29 = 61 miles.

 

On the northbound portion of the drive, 25 of

the 32 miles are within the range of the transmitter.

 

Now the question remains if any of the line

y=32 lies within the radius of the circle

for x= 0 to 29.

 

It is easy to see numerically that it does it.

Algebraically,

 

x^2 + 32^2 = 25^2

 

x^2 + 1024 = 625

 

x^2 = -399

 

The square of the number cannot be negative.

If solved for x, by taking the square root

of both sides, the value for x shall be imaginary,

namely square-root (399) * i

 

So only 25 of the 61 miles were within the range of

the transmitter.

 

25/61 = .409836 < 41%