Circles

Non-intersecting line (with respect to a circle)

A line in the plane of a circle that has no point in common with the circle. The other two possibilities are a secant, meeting it at two points, and a tangent, meeting it at exactly one.

Secant

A line that meets a circle at two distinct points. A tangent can be seen as the limiting case of a secant whose two meeting points have moved together into one.

Tangent to a circle

A line that meets a circle at exactly one point. The word comes from the Latin 'tangere', to touch, and there is only one tangent at any point of a circle.

Point of contact

The single point shared by a circle and one of its tangents; we say the tangent touches the circle there.

Parallel tangents: how many tangents to a circle can be parallel to a given secant?

At most two, one on each side of the secant. Moving a parallel line away from the centre shortens the chord it cuts until it touches the circle, which happens once on each side.

Tangent and radius theorem

At every point of a circle, the tangent and the radius drawn to that point meet at right angles.

How is it proved that the tangent at P is perpendicular to the radius OP?

Take any other point Q on the tangent. Q must lie outside the circle, otherwise the line would be a secant, so OQ > OP. Hence OP is the shortest distance from O to the tangent, and the shortest segment from a point to a line is perpendicular to it.

How many tangents to a circle can be drawn through a point inside it, on it, and outside it?

None through a point inside, since every line through it cuts the circle twice. Exactly one through a point on the circle. Exactly two through a point outside the circle.

Length of a tangent

The length of the segment of a tangent from an external point to its point of contact with the circle.

Equal tangents theorem

From any point outside a circle, the two tangents to the circle have equal lengths, measured from the point to the points of contact.

How is it proved that the two tangents PQ and PR from an external point P are equal?

Join OP, OQ and OR. Angles OQP and ORP are right angles by the tangent-radius theorem. The right triangles share the hypotenuse OP and have OQ = OR as radii, so they are congruent by RHS, giving PQ = PR.

Tangent length, radius and distance from the centre

If a tangent from P touches the circle at Q, then PQ² = OP² − OQ², i.e. (tangent length)² = (distance of P from the centre)² − (radius)². It is Pythagoras theorem in the right triangle OQP.

Why does the centre of a circle lie on the bisector of the angle between two tangents from an external point?

The congruent right triangles OQP and ORP give ∠OPQ = ∠OPR. So OP splits the angle between the tangents into two equal parts, and the centre O lies on its bisector.

In two concentric circles, how do you show that a chord of the larger circle which touches the smaller circle is bisected at the point of contact?

Join the centre O to the point of contact P. The chord is a tangent to the smaller circle at P, so OP is perpendicular to it. OP is also a perpendicular from the centre of the larger circle to its chord, and such a perpendicular bisects the chord.

Tangents TP and TQ are drawn from T to a circle with centre O. How do you show ∠PTQ = 2∠OPQ?

Let ∠PTQ = θ. TP = TQ, so triangle TPQ is isosceles and ∠TPQ = 90° − θ/2. The radius makes ∠OPT = 90°, so ∠OPQ = 90° − (90° − θ/2) = θ/2, which gives ∠PTQ = 2∠OPQ.

PQ is an 8 cm chord of a circle of radius 5 cm, and the tangents at P and Q meet at T. How do you find TP?

OT is perpendicular to PQ and bisects it at R, so PR = 4 cm and OR = √(5² − 4²) = 3 cm. Triangles TRP and PRO are similar by AA, so TP/PO = RP/RO, giving TP = 5 × 4/3 = 20/3 cm.

Why is the angle between two tangents from an external point supplementary to the angle their points of contact subtend at the centre?

The two radii to the points of contact are perpendicular to the tangents, so the quadrilateral formed by the centre, the two contact points and the external point has two right angles. Its angles total 360°, so the remaining two angles add to 180°.

XY and X′Y′ are parallel tangents, and a third tangent meets them at A and B. How do you prove ∠AOB = 90°?

From each of A and B two tangents are drawn, so OA bisects the angle at A and OB bisects the angle at B. Those two full angles are co-interior angles between the parallel tangents, adding to 180°. Their halves add to 90°, so in triangle AOB the angle at O is 90°.

If a quadrilateral ABCD circumscribes a circle, why is AB + CD = AD + BC?

Each side is made of two tangent segments, one from each end vertex. The two tangents from any single vertex are equal, so AB + CD and AD + BC each use exactly one tangent length from every vertex, and the sums match.

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