# Circles & Vectors

• Oct 16th 2008, 05:34 AM
DJ Hobo
Circles & Vectors
In circle geometry, there is a mathematical property that states a radius, drawn to bisect a chord, will meet the chord at 90 degrees.

I must prove this property using vector methods. How do I do this? I posted a similar question here some times ago that uses non-vector methods. I did eventually figure that out, but now I need to use vector methods.

Any help would be appreciated.
• Oct 16th 2008, 12:00 PM
earboth
Quote:

Originally Posted by DJ Hobo
In circle geometry, there is a mathematical property that states a radius, drawn to bisect a chord, will meet the chord at 90 degrees.

I must prove this property using vector methods. How do I do this? I posted a similar question here some times ago that uses non-vector methods. I did eventually figure that out, but now I need to use vector methods.

Any help would be appreciated.

1. Let C denote the center of the circle and the origin.

Then the vectors $\vec a$ and $\vec b$ are position vectors pointing at points on the circle line. Therefore

$|\vec a| = |\vec b| = r$

2. The vector $\overrightarrow{AB} = \vec b - \vec a$

3. The vector $\vec m = \frac12(\vec a + \vec b)$ has the same direction as the line passing through C and M, the midpoint of $\overline{AB}$

4. Calculate

$\vec m \cdot \overrightarrow{AB} = \frac12(\vec a + \vec b) \cdot (\vec b - \vec a) = \frac12\left( \vec a \vec b + (\vec b)^2 - (\vec a)^2 - \vec a \vec b \right)$

Since $|\vec a| = |\vec b|$ the value in the bracket is zero. Therefore $\vec m$ and $\overrightarrow{AB}$ are perpendicular.