Circle theorems II

Why circles have these properties Extended

A circle is symmetrical about every diameter: fold it along any line through the centre and the two halves match. All three properties in this lesson come from that symmetry together with one fact you already know, that all radii are equal.

In every question, look for the right-angled triangle that the symmetry creates. It almost always has:

• a radius as its hypotenuse (or the line from the centre to an outside point);
• half a chord, or a tangent, as one side;
• the distance from the centre as the other side.

Then Pythagoras or trigonometry finishes the job. When you give a reason, use the property in the words of the syllabus: "the perpendicular bisector of a chord passes through the centre", "equal chords are equidistant from the centre", "tangents from an external point are equal in length".

The perpendicular bisector of a chord Extended

The perpendicular bisector of a chord passes through the centre.

Why it is true. The centre $O$ is the same distance from $A$ as from $B$, because $OA$ and $OB$ are radii. The perpendicular bisector of $AB$ is exactly the set of points that are the same distance from $A$ and from $B$ (it is the locus of points equidistant from $A$ and $B$). So $O$ must be on it.

The same fact can be read in two other ways, and you may use either:

• the line from the centre to the midpoint of a chord is perpendicular to the chord;
• the perpendicular from the centre to a chord bisects the chord.

Using it to find a length. If a chord of length $2h$ is at distance $d$ from the centre of a circle of radius $r$, the half-chord, the distance and the radius make a right-angled triangle, so $h^2+d^2=r^2$. Always halve the chord first.

Using it to find the centre. Draw the perpendicular bisectors of two different chords. Each passes through the centre, so the centre is where they cross. On a grid, a horizontal chord from $x=-1$ to $x=5$ has perpendicular bisector $x=2$; a vertical chord from $y=-2$ to $y=4$ has perpendicular bisector $y=1$; the centre is $(2,1)$.

Equal chords are equidistant from the centre Extended

Equal chords are equidistant from the centre. If two chords of the same circle have the same length, their perpendicular distances from the centre are equal. The reverse is also true: chords the same distance from the centre are the same length.

Why it is true. For each chord, the half-chord $h$, the distance $d$ and the radius $r$ satisfy $h^2+d^2=r^2$. The radius is the same for both chords, so if the chords are equal (same $h$), then $d^2$ is the same, and so is $d$. Another way to see it: you can rotate the circle about $O$ to move one chord exactly onto the other.

The distance from the centre to a chord always means the perpendicular distance, $OM$, where $M$ is the midpoint. It is not the distance to an end of the chord, which is just the radius.

The same formula tells you that the closer a chord is to the centre, the longer it is. The longest chord of all is a diameter, at distance 0.

Tangents from an external point Extended

Tangents from an external point are equal in length. From a point $T$ outside a circle you can draw exactly two tangents, touching the circle at $A$ and $B$. Then $TA=TB$.

Why it is true. Join $OA$, $OB$ and $OT$. The angles at $A$ and $B$ are both ${90}^{°}$ (tangent and radius). The triangles $OAT$ and $OBT$ share the hypotenuse $OT$ and have $OA=OB$ (radii), so they are congruent (RHS), and $TA=TB$. By Pythagoras: $T{A}^2=O{T}^2-r^2=T{B}^2$.

The congruent triangles give more than equal tangents:

• $OT$ is a line of symmetry of the whole figure;
• $OT$ bisects angle $ATB$ and angle $AOB$;
• $OATB$ is a kite, with right angles at $A$ and $B$, so angle $AOB$ + angle $ATB={180}^{°}$;
• triangle $TAB$ is isosceles, so its base angles are equal.

Circles inside polygons. When a circle touches every side of a triangle or quadrilateral, each vertex is an external point with two equal tangents. Label the tangent lengths from each vertex with one letter, and the sides become sums: in a quadrilateral, this gives $AB+CD=BC+DA$.