HG10 - Tangents and chords

Radii and tangents

A tangent looks as though it just brushes a circle. What exact angle does it make with the radius, and why must two tangents from one point have equal lengths?

A tangent is a straight line that meets the circle at exactly one point, its point of contact. This is a stated geometric condition, not something to infer from an almost-touching drawing.

Proof. Let the radius be $r$. Suppose $OT$ were not perpendicular to the tangent. Drop a perpendicular from $O$ to the line at $H$. The perpendicular is shorter than the sloping segment $OT$, so $OH<r$: the line passes through a point inside the circle. It would then cut the circle twice, contradicting tangency. More explicitly, the two points on the line at distances $\sqrt{r^2-OH^2}$ on either side of $H$ both have distance $r$ from $O$, by Pythagoras. Therefore the supposed non-perpendicular tangent cannot exist.

Now take an external point $P$ and its two tangents to the same circle, meeting it at distinct points $A,B$. Join $OA,OB,OP$. Triangles $OAP$ and $OBP$ have right angles at the contact points, equal radii $OA=OB$ and common hypotenuse $OP$. They are congruent by RHS. Hence

Their corresponding angles at $P$ are also equal: $OP$ bisects the angle between the tangents. Equal tangent lengths refer to segments from the same external point to the same circle.

Keep the tangent length but change the angle between the tangents. The tangent equality remains valid while the central angle changes.

The contact points locate both right angles; the circle centre need not lie on either tangent.

[DIAGRAM: asset_name: Two tangents from one external point; asset_slug: equal-tangents-pair; caption: Equal tangent segments and two contact right angles link the external and central angles.; description: Exact circle panels construct tangents PA and PB from one external point P. Radii OA and OB meet them at right angles, OP is drawn, and the angles APB68 and92 produce central angles112 and88. Matching marks identify PA=PB.]

Use fresh tangent data and name the reasons.

Chord geometry

A radius drawn to a random point of a chord does not necessarily meet it at a right angle. A perpendicular from the centre, however, finds the chord's midpoint. The reverse statement is useful too.

Proof from the midpoint. Let $A,B$ be distinct endpoints of a non-diameter chord, $O$ the centre, and $M$ its midpoint. Join $OA,OB,OM$. Triangles $OMA,OMB$ have $OA=OB$ (radii), $AM=MB$ (midpoint), and common side $OM$. By SSS they are congruent. Thus $\angle OMA=\angle OMB$; they also sum to $180^\circ$ on the straight line $AB$. Each is $90^\circ$.

Proof from the perpendicular. If $OM$ meets chord $AB$ at a right angle, triangles $OMA,OMB$ have equal hypotenuses $OA,OB$ and common side $OM$. RHS congruence gives $AM=MB$, so $M$ is the midpoint. If the chord is a diameter, its midpoint is the centre itself and its two halves are radii; this case does not require a degenerate triangle proof. The perpendicular bisector still passes through that midpoint.

For calculations, let the radius be $r$, the perpendicular distance from the centre to the chord be $d$, and the half-chord length be $h$. The right triangle has hypotenuse $r$, so

Here $0\le d<r$ for a genuine chord with distinct endpoints. At $d=0$ the chord is a diameter. At $d=r$ the intersection has collapsed to a single contact point, rather than a two-endpoint chord.

Keep the radius but move the chord farther from the centre. Its length decreases.