Chords and the centre of the circle

Theorem: A perpendicular to a chord through the centre of the circle bisects the chord.

Given: $\displaystyle O$ is the centre of the circle, $\displaystyle AB$ a chord, $\displaystyle OM\perp AB$ ($\displaystyle M$ on $\displaystyle AB$).
To prove: $\displaystyle {AM}={BM}$.

The proof below draws the radii $\displaystyle OA$ and $\displaystyle OB$ and shows that the triangles $\displaystyle OMA$ and $\displaystyle OMB$ are congruent:

The first theorem is about one line with two of three properties: it passes through the centre, it is perpendicular to the chord, it bisects the chord. Any two of them bring the third:

• Theorem: A line through the centre of the circle that bisects a chord is perpendicular to the chord. You will write the proof yourself, in Exercise 1.
• Theorem: A line perpendicular to a chord that bisects it passes through the centre of the circle. The reason: $\displaystyle {OA}={OB}$, so $\displaystyle O$ is on the perpendicular bisector of $\displaystyle AB$ – the locus of the points at equal distances from the ends of the segment.

From the last theorem: the centre of a circle is on the perpendicular bisector of every chord. So a centre is found where the perpendicular bisectors of two chords meet.

The distance of a chord from the centre is the length of the perpendicular from the centre to the chord, $\displaystyle {d}={OM}$. The perpendicular bisects the chord, so in the right-angled triangle $\displaystyle OMA$ the legs are $\displaystyle d$ and $\displaystyle \frac{AB}{2}$, and the hypotenuse is the radius.

By Pythagoras’ theorem:

Example: the radius of the circle is $\displaystyle 10$, and the chord is $\displaystyle 16$ long. How far is the chord from the centre?

On the axes, we check that the segment from the centre to the chord’s midpoint is perpendicular to the chord: the product of the slopes of perpendicular lines is $\displaystyle -1$.

Example: the circle $\displaystyle {(x-2)^{2}+(y-5)^{2}}={25}$, with centre $\displaystyle K(2,5)$.
$\displaystyle AC$ is a chord: $\displaystyle A(6,2)$ and $\displaystyle C(5,9)$.

• The chord’s midpoint: $\displaystyle {M\left(\frac{6+5}{2},\frac{2+9}{2}\right)}={(5.5,\ 5.5)}$.
• The chord’s slope: $\displaystyle {m_{AC}}={\frac{9-2}{5-6}}={-7}$.
• The slope of $\displaystyle KM$: $\displaystyle {m_{KM}}={\frac{5.5-5}{5.5-2}}={\frac{0.5}{3.5}}={\frac{1}{7}}$.
• $\displaystyle {-7\cdot\frac{1}{7}}={-1}$, so $\displaystyle KM\perp AC$ – as the theorem says.

$\displaystyle O$ is the centre of the circle, and $\displaystyle M$ is the midpoint of the chord $\displaystyle AB$.
Prove: $\displaystyle OM\perp AB$.

The radius of the circle is $\displaystyle 13$, and the chord is $\displaystyle 24$ long.
How far is the chord from the centre of the circle?

The radius of the circle is $\displaystyle 6$, and the chord $\displaystyle AB$ is $\displaystyle 4$ from the centre.
How long is the chord? Leave the answer with a root, in its simplest form.

$\displaystyle O$ is the centre of the circle, and $\displaystyle M$ is the midpoint of the chord $\displaystyle AB$.
$\displaystyle OM$ extended meets the circle at $\displaystyle C$.
Prove: $\displaystyle {CA}={CB}$.

The circle: $\displaystyle {x^{2}+y^{2}}={25}$, with centre $\displaystyle O(0,0)$.
$\displaystyle AB$ is a chord: $\displaystyle A(-3,4)$ and $\displaystyle B(5,0)$.
$\displaystyle M$ is the chord’s midpoint.

Find the coordinates of $\displaystyle M$.