Circles and line segments

In this section, we will discuss how to work with circles and line segments. Specifically, we will look at how to find the center of a circle when given the two endpoints of a diameter, how to find the other endpoint of the diameter using the center, and how to find the equation of a perpendicular line to the diameter when only the endpoints are given.

1. Finding the Centre of a Circle Given Two Endpoints of the Diameter

The center of a circle is located at the midpoint of its diameter. If you are given two endpoints of the diameter, you can find the coordinates of the center by calculating the midpoint of the two points.

Let the endpoints of the diameter be \( (x_1, y_1) \) and \( (x_2, y_2) \). The midpoint (which is also the center of the circle) is found using the midpoint formula:

\[ \text{Centre} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

For example, if the endpoints of the diameter are \( (1, 3) \) and \( (5, 7) \), the center of the circle would be:

\[ \text{Centre} = \left( \frac{1 + 5}{2}, \frac{3 + 7}{2} \right) = (3, 5) \]

2. Finding the Other Endpoint of the Diameter Using the Centre

Once you have the center of the circle, you can use it to find the other endpoint of the diameter. If you know the center \( (x_c, y_c) \) and one endpoint of the diameter \( (x_1, y_1) \), you can use the midpoint formula in reverse to find the coordinates of the second endpoint \( (x_2, y_2) \).

The reverse midpoint formula is:

\[ x_2 = 2x_c - x_1 \quad \text{and} \quad y_2 = 2y_c - y_1 \]

For example, if the center of the circle is \( (3, 5) \) and one endpoint of the diameter is \( (1, 3) \), the other endpoint can be found as follows:

\[ x_2 = 2 \times 3 - 1 = 5 \quad \text{and} \quad y_2 = 2 \times 5 - 3 = 7 \]

Thus, the other endpoint of the diameter is \( (5, 7) \).

3. Finding the Equation of a Perpendicular Line to the Diameter Given the Endpoints

A line perpendicular to the diameter of a circle will pass through the center of the circle. The slope of the perpendicular line is the negative reciprocal of the slope of the diameter. Here’s how to find the equation of the perpendicular line:

1. Find the slope of the diameter: The slope of the line passing through the endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
2. Find the slope of the perpendicular line: The slope of the perpendicular line is the negative reciprocal of the slope of the diameter:
3. Write the equation of the perpendicular line: Use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the center of the circle and \( m \) is the slope of the perpendicular line. You can now plug in the values to find the equation.

For example, suppose the endpoints of the diameter are \( (1, 3) \) and \( (5, 7) \), and we have already found the center \( (3, 5) \). The slope of the diameter is:

\[ \text{slope of diameter} = \frac{7 - 3}{5 - 1} = \frac{4}{4} = 1 \]

Thus, the slope of the perpendicular line is:

\[ \text{slope of perpendicular line} = -\frac{1}{1} = -1 \]

Now, using the point-slope form with the center \( (3, 5) \) and slope \( -1 \), the equation of the perpendicular line is:

\[ y - 5 = -1(x - 3) \]

Simplifying the equation:

\[ y - 5 = -x + 3 \] \[ y = -x + 8 \]

Thus, the equation of the perpendicular line is \( y = -x + 8 \).

4. Summary:

• The center of a circle can be found by calculating the midpoint of the two endpoints of the diameter using the formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
• Once the center is known, the other endpoint of the diameter can be found by using the reverse midpoint formula: \( x_2 = 2x_c - x_1 \) and \( y_2 = 2y_c - y_1 \).
• The equation of the perpendicular line to the diameter can be found by first calculating the slope of the diameter, then finding the negative reciprocal for the perpendicular slope, and using the point-slope form to write the equation of the line passing through the center of the circle.

These methods are useful for solving problems involving circles, line segments, and perpendicular lines, and they form the basis for many geometric applications.