A small radio transmitter broadcasts in a 25 mile radius. If you drive along a straight line from a city 30 miles north of the transmitter to a second city 40 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

Since, the radio transmitter broadcasts the signal in a 25 mile radius. Therefore, the effective range of the broadcast will be in a shape of circle. And the radius of this circular region will be 25 miles.

Now, let the centre of this circular region be located at the point (0, 0). Thus, the equation of this effective circular region will be obtained as:

We know that the equation of a circle in the standard form is as:

(x – h)² + (y – k)² = r²

Here, r is the length of radius of the circle and (h, k) is the coordinate of the centre of the circle. Now, putting (h, k) = (0, 0) and r = 25 in the above formula, we get

Or, (x – 0)² + (y – 0)² = 25²

Or, x² + y² = 625     …… Equation (1)

Now, the coordinate of the first city which is 30 miles in the North of the transmitter will be (0, 30). And the coordinate of the second city which is 40 miles in the east of the transmitter will be (40, 0).

Now, we will find the equation of the line which joins the points (0, 30) and (40, 0). The formula to find the equation of a line which passes through two points is as:

(y – y₁) = {(y₂ – y₁)/(x₂ – x₁)}(x – x₁)

Here, (x₁, y₁) and (x₂, y₂) are the two points. Putting (x₁, y₁) = (0, 30) and (x₂, y₂) = (40, 0) in the above formula, we get

Or, (y – 30) = {(0 – 30)/(40 – 0)}(x – 0)

Or, (y – 30) = – 30x/40

Or, y = – 3x/4 + 30 …… Equation (2)

Putting y = – 3x/4 + 30 in Equation (1), we get

x² + y² = 625

Or, x² + (– 3x/4 + 30)² = 625

Or, x² + (30 – 3x/4)² = 625

Or, x² + (30)² + (3x/4)² – 2 × 30 × 3x/4 = 625

Or, x² + 900 + 9x²/16 – 45x = 625

Or, x² + 900 + 9x²/16 – 45x – 625 = 0

Or, x² + 9x²/16 – 45x + 900 – 625 = 0

Or, x² + 9x²/16 – 45x + 275 = 0

Or, (16x² + 9x² – 16 × 45x + 16 × 275)/16 = 0

Or, 25x² – 720x + 4400 = 0

Or, 5(5x² – 144x + 880) = 0

Or, 5x² – 144x + 880 = 0

Or, 5x² – (100 + 44)x + 880 = 0

Or, 5x² – (100x + 44x) + 880 = 0

Or, 5x² – 100x – 44x + 880 = 0

Or, 5x(x – 20) – 44(x – 20) = 0

Or, (x – 20)(5x – 44) = 0

Either (x – 20) = 0 or (5x – 44) = 0

x – 20 = 0

Or, x = 20

5x – 44 = 0

Or, 5x = 44

Or, x = 44/5

Or, x = 8.8 Thus, the values of x are 20 and 8.8. Now, putting these value of x in equation (2), we get

Putting x = 20 in equation (2).

y = – (3 × 20)/4 + 30

y = – 15 + 30

y = 15

Putting x = 8.8 in equation (2).

y = – (3 × 8.8)/4 + 30

y = – 6.6 + 30

y = 23.4

Thus, the signal will receive between the points (20, 15) and (8.8, 23.4). Now, the distance between these two points can be obtained as:

The distance between the points = {(x₂ – x₁)² + (y₂ – y₁)²}^1/2

Here, (x₁, y₁) and (x₂, y₂) are the two points. Putting (x₁, y₁) = (20, 15) and (x₂, y₂) = (8.8, 23.4) in the above formula, we get

The distance between the points

= {(8.8 – 20)² + (23.4 – 15)²}^1/2

= {(- 11.2)² + (8.4)²}^1/2

= {125.44 + 70.56}^1/2

= {196}^1/2

= 14 miles

Now, we will have to find the distance between the points (0, 30) and (40, 0). The distance between these two points can be obtained as:

The distance between the points = {(x₂ – x₁)² + (y₂ – y₁)²}^1/2

Here, (x₁, y₁) and (x₂, y₂) are the two points. Putting (x₁, y₁) = (0, 30) and (x₂, y₂) = (40, 0) in the above formula, we get

The distance between the points

= {(40 – 0)² + (0 – 30)²}^1/2

= {(40)² + (- 30)²}^1/2

= {1600 + 900}^1/2

= {2500}^1/2

= 50 miles

Now, the percentage of signal received

= (14/50)×100%

= 28%

Hence, 28% of the drive will pick up the signal from the transmitter.