8.05e–8.05g - Chords and tangents

Bisecting a chord

A straight cut $12\,\mathrm{cm}$ long crosses a circular disc of radius $10\,\mathrm{cm}$. How far is the cut from the centre? Drawing the perpendicular from the centre splits the cut into two equal lengths and makes a right-angled triangle.

“If and only if” means both directions need reasons. The diagram shows the two triangles to compare.

[DIAGRAM: asset_name: A chord and its perpendicular bisector; asset_slug: chord-bisector-proof; caption: Equal radii make the two triangles congruent in either direction.; description: A circle with centre O above a horizontal non-diameter chord AB and midpoint M. Radii OA and OB and segment OM are drawn. AM and MB have matching ticks and a right-angle square marks the perpendicular at M. A second panel shows radius 10 cm, half-chord 6 cm and perpendicular distance 8 cm.]

Perpendicular implies bisected. If $OM\perp AB$, triangles $OMA$ and $OMB$ are right-angled, have equal hypotenuses $OA=OB$ and share side $OM$. They are congruent by RHS, so $AM=MB$.

Bisected implies perpendicular. If $AM=MB$, then $OA=OB$, $OM$ is shared and $AM=MB$. The triangles are congruent by SSS. Thus $\angle OMA=\angle OMB$; they lie on a straight line, so each is $90^\circ$.

The non-diameter condition avoids a collapsed triangle. If $AB$ is itself a diameter, its midpoint is $O$. Every line through $O$ then passes through its midpoint, so passage through the midpoint alone does not force a right angle. Its perpendicular bisector is still the particular line through $O$ perpendicular to $AB$.

Change only the chord length. A longer chord of the same circle sits closer to the centre.

Pythagoras is on the 2027 formula sheet; the chord-bisector theorem is not. The key decision is to halve the chord before substituting.

The radius and tangent

A tangent has one point in common with a circle, and the radius to that contact point meets it at a right angle.

Why must it be perpendicular? Drop a perpendicular from $O$ to the tangent line at $H$. Suppose $H\ne P$. Reflect $P$ in $H$ to get a second point $P'$ on the line. If $H=O$, reflection in $O$ gives $OP'=OP$ directly. Otherwise the two right-angled triangles $OHP$ and $OHP'$ have the same two perpendicular side lengths, so again $OP'=OP$. Thus $P'$ is also on the circle, contradicting the single contact point. Therefore $H=P$.

[DIAGRAM: asset_name: Why a tangent must be perpendicular to the radius; asset_slug: radius-tangent-proof; caption: A non-perpendicular line through P would also contain another circle point P prime.; description: Left panel shows a true tangent at P with radius OP and right-angle square. Right panel shows a crossing line with perpendicular foot H, distinct equal-distance intersections P and P prime, and the two radii. The crossing line is explicitly labelled not a tangent.]

Conversely, if a line through $P$ is perpendicular to radius $OP$, every other point $X$ on it satisfies $OX^2=OP^2+PX^2>OP^2$. Every such $X$ is outside the circle, so the line is a tangent. These arguments use distance and Pythagoras, rather than assuming the tangent theorem to prove itself.