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!