A small radio transmitter broadcasts in a 56 mile radius. If you drive along a straight line from a city 64 miles north of the transmitter to a second city 63 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

The problem involves a small radio transmitter with a broadcasting range of 56 miles, and we are to determine how much of a drive between two cities is within this range. Let’s break it down step-by-step:

### Step 1: Coordinates setup
- Let the location of the transmitter be at the origin \((0, 0)\).
- The first city is 64 miles north of the transmitter, so its coordinates are \((0, 64)\).
- The second city is 63 miles east of the transmitter, so its coordinates are \((63, 0)\).

We want to find out how much of the straight-line drive from the city \((0, 64)\) to \((63, 0)\) is within the 56-mile radius of the transmitter.

### Step 2: Equation of the path (line)
The line between the two cities is a straight line, and we can find its equation. The slope of the line is:

\[
\text{slope} = \frac{0 - 64}{63 - 0} = -\frac{64}{63}
\]

Thus, the equation of the line in point-slope form is:

\[
y - 64 = -\frac{64}{63}(x - 0)
\]

Simplifying:

\[
y = -\frac{64}{63}x + 64
\]

### Step 3: Equation of the circle (broadcasting range)
The transmitter's broadcasting range is a circle centered at the origin with a radius of 56 miles. The equation of this circle is:

\[
x^2 + y^2 = 56^2
\]

\[
x^2 + y^2 = 3136
\]

### Step 4: Points of intersection
We want to find the points on the line that are within the 56-mile radius, so we substitute the equation of the line \(y = -\frac{64}{63}x + 64\) into the circle equation:

\[
x^2 + \left(-\frac{64}{63}x + 64\right)^2 = 3136
\]

Now, expand the quadratic term:

\[
x^2 + \left(\frac{4096}{3969}x^2 - \frac{8192}{63}x + 4096\right) = 3136
\]

Multiply everything by 3969 to eliminate the fractions:

\[
3969x^2 + 4096x^2 - 8192 \cdot 63x + 4096 \cdot 3969 = 3136 \cdot 3969
\]

We can now solve this quadratic equation for \(x\), which will give us the \(x\)-coordinates of the intersection points. Once we have these, we can calculate the total distance between these two points, which represents how much of the drive will be within the broadcast range.

I'll now solve for \(x\).The two \(x\)-coordinates where the line intersects the circle are:

\[
x_1 = \frac{258048}{8065} - \frac{3528 \sqrt{2881}}{8065}
\]
\[
x_2 = \frac{258048}{8065} + \frac{3528 \sqrt{2881}}{8065}
\]

These values represent the points of intersection along the x-axis. Now, to find the distance between these two points, which is the part of the path within the broadcast range, we will calculate \(x_2 - x_1\).

I'll proceed to calculate that distance.The distance during which the drive will be within the 56-mile broadcast range of the transmitter is approximately **46.96 miles**.

Would you like further details on any of the steps? Here are some related questions to expand on this topic:

1. How do you find the distance between two points on a line?
2. What is the equation of a circle, and how is it derived?
3. How do you solve systems of equations involving circles and lines?
4. What are some real-world applications of intersection points between curves and lines?
5. How can the slope of a line be used in different geometrical problems?

**Tip:** When dealing with word problems, always try to visualize the scenario using sketches to better understand the relationships between elements.

Calculate how much of a straight-line drive from a city 64 miles north to another city 63 miles east falls within a 56-mile broadcast radius of a transmitter. This problem involves geometry concepts like the equation of a circle, line, and their intersection points. Suitable for students in Grades 10-12.

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