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Circle theorems connect radii, chords, tangents and secants. Their usefulness comes from a small set of relationships that can be justified through congruent or similar triangles. The fixed radius determines relationships among the chords, tangents and secants that meet the circle.

This page focuses on length relationships and their proofs. Angles in Circles develops the angle-and-arc rules. The centre, radius and equation of a coordinate circle are treated in Circles in Coordinate Geometry.

On this page

The vocabulary identifies the configuration

A radius joins the centre to a point on the circle. A chord joins two points on the circle. A diameter is a chord through the centre. A secant is a line meeting the circle at two distinct points, while a tangent meets it at one point.

The same infinite line may have an external segment and a segment inside the circle. When a theorem uses a whole secant length, that length begins at the external point and ends at the far circle intersection. It is not just the part inside the circle.

A perpendicular from the centre bisects a chord

Let ABAB be a non-diameter chord and let the perpendicular from centre OO meet it at MM. The right triangles OMAOMA and OMBOMB have equal hypotenuses OA=OBOA=OB, because both are radii, and the common leg OMOM. RHS congruence gives AM=MBAM=MB.

Conversely, if MM is the midpoint of a non-diameter chord, then OMOM is perpendicular to the chord. For a diameter, the centre is already its midpoint because the two halves are radii. The centre and midpoint then coincide, so they do not determine a unique line.

A perpendicular radius bisects a chord

A perpendicular radius bisects a chordA chord lies three units from the centre of a radius-five circle. Its midpoint splits it into two four-unit halves.OABM3544
OM is perpendicular to AB; the right triangle OMB gives MB = √(25 − 9) = 4.
Diagram explanation and exact data

A=(−4,3), B=(4,3) and O=(0,0). Both radii have length 5. RHS congruence of OMA and OMB proves the chord-bisection relationship without reading lengths from the drawing.

A perpendicular radius bisects a chord: exact reference data
Measure Value
Radius 55
Centre-to-chord distance 33
Half-chord 44
Chord 88

Finding a chord length

A circle has radius 5 cm5\text{ cm}, and a chord is 3 cm3\text{ cm} from the centre. Half the chord is a leg of a right triangle:

AM=52−32=4 cm,AB=8 cm.AM=\sqrt{5^2-3^2}=4\text{ cm},\qquad AB=8\text{ cm}.

The common error is to stop at 4 cm4\text{ cm}, which gives only half the required length. The chord lies inside a diameter of length 10 cm10\text{ cm}, so an 8 cm8\text{ cm} result is plausible.

In the same circle, or in equal-radius circles, equal chords have equal perpendicular distances from the centre, and conversely. A chord closer to the centre is longer. The same-radius condition matters: equal lengths in different-sized circles do not determine equal central distances.

A tangent is perpendicular to its radius

At a point of tangency TT, the radius OTOT is perpendicular to the tangent line. Another line through TT that is not perpendicular passes through the circle’s interior and meets the circle again. The perpendicular therefore identifies the tangent direction.

If an external point PP is 13 cm13\text{ cm} from the centre of a radius-5 cm5\text{ cm} circle, a tangent segment PTPT has length

PT=OP2−OT2=132−52=12 cm.PT=\sqrt{OP^2-OT^2}=\sqrt{13^2-5^2}=12\text{ cm}.

The right angle is at TT, so OPOP is the hypotenuse. It is not correct to label the tangent segment as the hypotenuse just because it is outside the circle.

Tangents from one external point are equal

Suppose PAPA and PBPB are tangents from the same external point PP to one circle. The triangles OAPOAP and OBPOBP are right-angled, have common hypotenuse OPOP and equal radius legs OA=OBOA=OB. They are congruent by RHS, so

PA=PB.PA=PB.

If the lengths are 3x−23x-2 and x+6x+6, then 3x−2=x+63x-2=x+6 gives x=4x=4. Each tangent segment is 1010 units. Tangents drawn from two different external points do not generally have equal lengths.

Intersecting chords

If chords ABAB and CDCD intersect at an interior point PP, then

PA⋅PB=PC⋅PD.PA\cdot PB=PC\cdot PD.

To see why, compare △APC\triangle APC and △DPB\triangle DPB. Their angles at PP are vertically opposite. Their other matching angles are inscribed angles subtending the same arcs. The triangles are similar, so PA/PD=PC/PBPA/PD=PC/PB, which gives the product equation.

If PA=3PA=3, PB=8PB=8 and PC=4PC=4, then 3(8)=4(PD)3(8)=4(PD) and PD=6PD=6. Each product uses the two pieces of a single chord, not adjacent segments chosen by appearance.

Secants and the tangent-secant theorem

Let PP be outside the circle. Suppose one secant reaches the near point AA and then the far point BB, while another reaches CC then DD. The relationship is

PA⋅PB=PC⋅PD.PA\cdot PB=PC\cdot PD.

The symbols look like the intersecting-chords equation, but the lengths now mean external part times whole secant. If PA=4PA=4 and AB=5AB=5, the whole length is PB=9PB=9. If PC=3PC=3, then 3(PD)=4(9)=363(PD)=4(9)=36, so PD=12PD=12 and the interior part CD=9CD=9.

For a tangent PTPT and a secant through A,BA,B from the same external point,

PT2=PA⋅PB.PT^2=PA\cdot PB.

With external part 44 and interior part 55, the tangent length is 4(9)=6\sqrt{4(9)}=6. This statement concerns segment lengths, not angles. The AMSI circle module develops these relationships through similarity.

Choosing a theorem from the givens

First identify whether the known point lies inside, on or outside the circle. Next label the complete lines and the segment endpoints. For a chord-distance problem, draw the perpendicular radius to expose a right triangle. For two external lines, mark the near and far intersections before forming a product.

Treat the sketch as a record of the hypotheses. An apparently tangent line needs a stated tangency condition or a proof that it is perpendicular to the radius at a boundary point. A diagram touching the curve on screen is not itself evidence.

Practice questions

  1. A radius-13 cm13\text{ cm} circle has a chord 5 cm5\text{ cm} from its centre. Find the chord length.
  2. Intersecting chords have segment lengths 2,122,12 and 3,x3,x. Find xx.
  3. A secant from an external point has external part 44 and whole length 99. Find the tangent length from that point.
Check your answers
  1. Half the chord is 132−52=12 cm\sqrt{13^2-5^2}=12\text{ cm}, so the whole chord is 24 cm24\text{ cm}.
  2. 2(12)=3x2(12)=3x, so x=8x=8.
  3. t2=4(9)=36t^2=4(9)=36, so the positive tangent length is t=6t=6. Do not add 44 to 99: the whole length was already supplied.

Related topics

Angles in Circles explains the inscribed-angle facts used in the similarity argument. Chords, Arcs and Sectors separates straight chord length from curved arc length and region area.

Sources and further reading