1. Find the longest and shortest distance from (10, 7) to the circle x^2 + y^2-4x-2y-20=0

2. From the start, a man runs 4km towards north, 3 km towards east and takes a rest. Draw the line from starring point to resting point and line from resting point to finish point. Find the angle between the lines assuming that east-west is parallel to x-axis.
3. And arch 18months high has the form of a parabola with a vertical axis. The length of a horizontal beam placed across the arch 8m from the top is 64m. Find the width of the arch at the bottom.

The circle is

(x-2)^2 + (y-1)^2 = 25

The shortest distance is along the line from the center (2,1) to (10,7). So, find the distance between those two points, and subtract 5, the radius of the circle. For the longest distance add 5, to the opposite end of the diameter.

The runner rests at the end of the hypotenuse of a 3-4-5 triangle. You give no indication of the finish point.

The parabola has the equation
y = 18-kx^2
Since the point (32,10) is on the parabola, the equation is

y = 18 - 1/128 x^2
The roots are ±48
so the width at the bottom is 96m

1. To find the longest and shortest distances from the point (10, 7) to the circle x^2 + y^2 - 4x - 2y - 20 = 0, we can use the distance formula.

First, let's rewrite the equation of the circle in its standard form by completing the square:

(x^2 - 4x) + (y^2 - 2y) = 20
(x^2 - 4x + 4) + (y^2 - 2y + 1) = 20 + 4 + 1
(x - 2)^2 + (y - 1)^2 = 25

The center of the circle is at (2, 1) and the radius is √25 = 5.

Now, we can calculate the distances from the point (10, 7) to the center of the circle (2, 1):

Shortest distance: This occurs when the line connecting the point (10, 7) and the center of the circle (2, 1) is perpendicular to the circle's circumference. We can take the distance as the difference between the radius and the distance between the point and the center:

Shortest distance = 5 - √((10 - 2)^2 + (7 - 1)^2) = 5 - √(64 + 36) = 5 - √100 = 5 - 10 = -5 (neglecting the negative value)

Longest distance: This occurs when the line connecting the point (10, 7) and the center of the circle (2, 1) passes through the center of the circle. We can take the distance as the sum of the radius and the distance between the point and the center:

Longest distance = 5 + √((10 - 2)^2 + (7 - 1)^2) = 5 + √(64 + 36) = 5 + √100 = 5 + 10 = 15

Therefore, the shortest distance is -5 (neglected) and the longest distance is 15 units.

2. To draw the lines and find the angle between them, let's visualize the scenario:

-----------------------------------------------------------
| | Finish Point
| X |
| |
| Resting Point |
| | Starting Point
-----------------------------------------------------------

Using the given information, we know that the starting point is at (0, 0), the man runs 4km towards the north (positive y-direction), and then 3km towards the east (positive x-direction).

Now, let's plot the points on a coordinate plane:

Starting Point (0, 0)
Resting Point (3, 4) [3km east and 4km north]
Finish Point (3, 0) [3km east]

Now, we can draw the line from the starting point to the resting point (0, 0) to (3, 4), and the line from the resting point to the finish point (3, 4) to (3, 0):

-----------------------------------------------------------
| X |
| | Finish Point
| | |
| | |
| X_______________ |
| Starting Point Resting Point
-----------------------------------------------------------

To find the angle between the lines, we need to use trigonometry. Since east-west is parallel to the x-axis, the angle between the lines can be calculated as the angle between the x-axis and the line connecting the resting point and the finish point.

We can use the formula:

tan(angle) = (y2 - y1) / (x2 - x1)

Plugging in the values:

tan(angle) = (0 - 4) / (3 - 3) = -4 / 0

The denominator is zero, which means that the line connecting the resting point and the finish point is perpendicular to the x-axis. Therefore, the angle between the lines is 90 degrees (or π/2 radians).

3. To find the width of the arch at the bottom, we need to use the information provided. Let's visualize the arch:

Arch
|
| X
| |
| |
| |_______
| |<----- 18 m -------->|
| | |
| X

The arch is in the form of a parabola with a vertical axis. The length of a horizontal beam placed across the arch, 8m from the top, is 64m.

From the information given, we can deduce that the width of the arch at the top (distance between the two ends of the horizontal beam) is 64m.

Now, let's solve for the width of the arch at the bottom. Since it is a parabola, we know that the width of the arch remains constant for any horizontal beam placed parallel to the x-axis.

Therefore, the width of the arch at the bottom is also 64m.