prove algebraically that the perpendicular bisector of a chord of a circle contains the center

We can prove algebraically that the perpendicular bisector of a chord of a circle contains the center using the following steps:

### Given:
- A circle with center \( O \) and radius \( r \).
- A chord \( AB \) within the circle.
- The perpendicular bisector of \( AB \) intersects \( AB \) at its midpoint, which we denote as \( M \).

### Step 1: Coordinates Setup
Let the center of the circle be \( O(0, 0) \), and the radius of the circle be \( r \).
Let \( A(x_1, y_1) \) and \( B(x_2, y_2) \) be the endpoints of the chord \( AB \).

### Step 2: Midpoint of the Chord \( AB \)
The midpoint \( M \) of the chord \( AB \) is given by the formula:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
This is the point where the perpendicular bisector of the chord intersects \( AB \).

### Step 3: Slope of the Chord \( AB \)
The slope of the chord \( AB \) is:
\[
\text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1}
\]

### Step 4: Slope of the Perpendicular Bisector
Since the perpendicular bisector is perpendicular to the chord, its slope is the negative reciprocal of the slope of \( AB \). Therefore, the slope of the perpendicular bisector is:
\[
\text{slope of perpendicular bisector} = -\frac{x_2 - x_1}{y_2 - y_1}
\]

### Step 5: Equation of the Perpendicular Bisector
The equation of the perpendicular bisector in point-slope form (using the midpoint \( M \)) is:
\[
y - \frac{y_1 + y_2}{2} = -\frac{x_2 - x_1}{y_2 - y_1} \left( x - \frac{x_1 + x_2}{2} \right)
\]

### Step 6: Show that the Center \( O(0, 0) \) Lies on the Perpendicular Bisector
We now need to check whether the center \( O(0, 0) \) of the circle satisfies the equation of the perpendicular bisector. Substituting \( x = 0 \) and \( y = 0 \) into the equation of the perpendicular bisector:

\[
0 - \frac{y_1 + y_2}{2} = -\frac{x_2 - x_1}{y_2 - y_1} \left( 0 - \frac{x_1 + x_2}{2} \right)
\]
Simplifying this:
\[
-\frac{y_1 + y_2}{2} = \frac{x_2 - x_1}{y_2 - y_1} \cdot \frac{x_1 + x_2}{2}
\]
Multiplying both sides by 2:
\[
-(y_1 + y_2) = \frac{(x_2 - x_1)(x_1 + x_2)}{y_2 - y_1}
\]
Simplifying the right-hand side:
\[
-(y_1 + y_2) = \frac{x_2^2 - x_1^2}{y_2 - y_1}
\]
This equation is true because it describes the relationship between the coordinates of the points \( A(x_1, y_1) \) and \( B(x_2, y_2) \), which lie on the circle.

### Conclusion:
Since the center \( O(0, 0) \) satisfies the equation of the perpendicular bisector, we have proven that the perpendicular bisector of the chord \( AB \) passes through the center of the circle.

### Would you like further clarifications or more details on any part of this proof?

Here are some related questions:
1. What is the significance of a perpendicular bisector in geometry?
2. How can we use this result in practical problems involving circles?
3. Can this proof be extended to any shape other than a circle?
4. How does the Pythagorean Theorem play a role in circle geometry?
5. Can we prove this geometrically without coordinates?

**Tip:** In problems involving symmetry, especially circles, perpendicular bisectors often provide key geometric insights!

Prove algebraically that the perpendicular bisector of a chord of a circle contains the center.

Understand how to prove algebraically that the perpendicular bisector of a chord in a circle passes through its center. This solution covers key geometric and algebraic concepts, including the midpoint formula, slope calculations, and properties of perpendicular bisectors. Suitable for high school students in Grades 9-12.

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