Bisection of Chords (AQA A-Level Mathematics): Revision Notes

1. Chord of a Circle:

• A chord is a line segment with both endpoints on the circle.
• If $\ A(x_1, y_1)$ and $\ B(x_2, y_2)$ are endpoints of the chord, the chord $\ AB$ can be analysed using coordinate geometry.

2. Midpoint of the Chord:

• The midpoint $\ M$ of the chord $\ AB$ is given by: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

3. Perpendicular Bisector:

• The perpendicular bisector of a chord is the line that is perpendicular to the chord and passes through its midpoint.
• The perpendicular bisector of any chord passes through the centre $\ O$ of the circle.

The perpendicular bisector of a chord in a circle passes through the centre of the circle. Explanation:

• Given a circle with centre $\ O(h, k)$ and radius $\ r ,$ any chord $\ AB$ in the circle can be analysed using the midpoint formula and the concept of slopes.
• The line passing through $\ M$ (the midpoint of $\ AB$) that is perpendicular to $\ AB$ will have a slope that is the negative reciprocal of the slope of $\ AB$. If $\ M(x_m, y_m)$ is the midpoint of the chord $\ AB ,$ then the perpendicular bisector will have a line equation that can be found using the point-slope form of a line.

Example:

Consider a circle with equation $\ x^2 + y^2 = 25$. Let $\ A(4, 3)$ and $\ B(-4, -3)$ be the endpoints of a chord.

Step 1: Find the Midpoint of the Chord $\ AB :$

Using the midpoint formula: $M = \left(\frac{4 + (-4)}{2}, \frac{3 + (-3)}{2}\right) = (0, 0)$ So, $M$ is the origin.

Step 2: Find the Slope of $\ AB$:

Slope of $\ AB$: $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 3}{-4 - 4} = \frac{-6}{-8} = \frac{3}{4}$

Step 3: Find the Equation of the Perpendicular Bisector:

The slope of the perpendicular bisector will be the negative reciprocal of $\ \frac{3}{4}$, which is $\ -\frac{4}{3} .$

Since $\ M(0, 0)$ is the midpoint: $y - 0 = -\frac{4}{3}(x - 0)$ Simplifying, the equation of the perpendicular bisector is: $y = -\frac{4}{3}x$

Step 4: Confirm the Perpendicular Bisector Passes Through the Centre:

Since the centre of the circle is at the origin $\ (0, 0) ,$ and the perpendicular bisector also passes through $\ (0, 0)$, it confirms the theorem.

Given a circle with equation $\ x^2 + y^2 = 36 ,$ find the equation of the perpendicular bisector of the chord with endpoints $\ A(6, 0)$ and $\ B(0, 6) .$

We are given the equation of a circle, $x^2 + y^2 = 36$ , and asked to find the equation of the perpendicular bisector of the chord with endpoints $A(6, 0)$ and $B(0, 6)$.

Step 1: Find the midpoint of $AB$

The formula for the midpoint $M$ of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is:

Here, the coordinates of $A$ are $(6, 0)$, and the coordinates of $B$ are $(0, 6)$. Therefore, the midpoint $M$ is:

Step 2: Find the slope of the line segment $AB$

The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

Using $A(6, 0)$ and $B(0, 6)$, the slope of line $AB$ is:

Step 3: Find the slope of the perpendicular bisector

The slope of the perpendicular bisector is the negative reciprocal of the slope of the original line. If the slope of line $AB$ is $-1$, the slope of the perpendicular bisector is:

Step 4: Use the point-slope form to find the equation of the perpendicular bisector

The point-slope form of a line is given by:

where $(x_1, y_1)$ is a point on the line and $m$ is the slope.

We already know the slope of the perpendicular bisector is $1$ and it passes through the midpoint $M(3, 3)$. Substituting these values into the point-slope form:

Simplifying this equation:

Final Answer:

The equation of the perpendicular bisector is $y = x$ .