Perpendicular from the centre to a chord

Theorem. The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord. (Reason: line from centre to midpt of chord)

Proof (examinable). Let $O$ be the centre and $M$ the midpoint of chord $AB$. Join $OA$, $OB$ and $OM$.

In $△OMA$ and $△OMB$:

• $OA=OB$ (radii)
• $AM=MB$ (given)
• $OM=OM$ (common)

$△OMA\equiv △OMB$ (SSS), so ${\hat{M}}_1={\hat{M}}_2$.

But ${\hat{M}}_1+{\hat{M}}_2=180°$ ($\angle$s on a str line), so ${\hat{M}}_1={\hat{M}}_2=90°$ and $OM\perp AB$.

Learn this proof with every reason. Only the theorem is examinable as a proof; its converse is not.

Result. The line drawn from the centre of a circle perpendicular to a chord bisects the chord. (Reason: line from centre $\perp$ to chord)

You use this as a result in riders. It follows from RHS congruence: $OA=OB$ (radii), $OM$ common and ${\hat{M}}_1={\hat{M}}_2=90°$, so $AM=MB$.

A third related fact: the perpendicular bisector of a chord passes through the centre. (Reason: perp bisector of chord) This is how you find the centre of a circle from two chords.

The radius, the distance from the centre to the chord, and half the chord form a right-angled triangle:

$r^2=d^2+{\left(\frac{\text{chord}}{2}\right)}^2$

Example. Radius 10 cm, distance from the centre 6 cm: half the chord is $\sqrt{100-36}=8$, so the chord is 16 cm.

Example: a pipe. Water in a pipe has a surface 48 cm wide and is 12 cm deep at the deepest point. If the radius is $r$, the centre is $r-12$ above the water surface:

$(r-12{)}^2+24^2=r^2\Rightarrow -24r+720=0\Rightarrow r=30\text{ cm}$

Draw the radius to the end of the chord. That is the hypotenuse.