0580 · 4.8Circle theorems II

Chord perpendicular bisectors

0:00A circle has no preferred direction, and that symmetry gives us useful information about lengths. We can locate its centre, compare chords, and calculate tangent lengths without measuring a diagram. In this video, we will explain why those relationships work. The key is to connect the circle’s symmetry with equal radii and matching triangles. A chord is a straight line segment joining two points on a circle. A radius joins the centre to a point on the circumference.

0:24All radii of the same circle have equal length. A tangent is a straight line that touches a circle at exactly one point. We also need the meaning of a perpendicular bisector. Perpendicular means meeting at ninety degrees, while bisector means dividing something into two equal parts. The perpendicular bisector of a segment passes through its midpoint at ninety degrees. Both conditions matter, because a line can divide a segment equally without being perpendicular.

0:51Join the centre O to the endpoints P and Q of a chord. The two new segments are radii, so triangle OPQ has two equal sides. Place M at the midpoint of the chord, then join O to M. This divides the triangle into two smaller triangles. The smaller triangles match because they have three corresponding sides equal. Their radii are equal, their chord halves are equal, and they share the side from O to M.

1:15Matching triangles have equal corresponding angles. Therefore, the two angles at M are equal. Those angles also lie along a straight line, so together they make one hundred and eighty degrees. Since they are equal, each angle is ninety degrees. The line through O and M is therefore the chord’s perpendicular bisector. This explains the property that the perpendicular bisector of a chord passes through the centre. We can also use this relationship when a perpendicular from the centre is already shown.

1:42The two right-angled triangles have equal hypotenuses because those sides are radii. They also share the perpendicular side, so the triangles match. Their chord segments are equal, meaning the perpendicular from the centre bisects the chord.

Distance from centre to chord

1:55Imagine finding the centre of a circular piece of card whose centre is unmarked. The perpendicular bisector of one chord gives a line containing the centre. A second suitable chord gives another such line. Their intersection locates the centre because both lines must pass through it. In our first example, a circle with centre O has a radius of ten centimetres. The chord from P to Q is sixteen centimetres long.

2:19The line from O meets the chord at M, at ninety degrees. We must calculate the distance from the centre to the chord and give a geometrical reason. The perpendicular from the centre to a chord bisects the chord, so M is its midpoint. Dividing sixteen by two gives eight centimetres for the length from M to Q. Triangle OMQ has a right angle at M, and its hypotenuse is the radius from O to Q.

2:43Pythagoras' theorem adds the squares of the two shorter sides to give the square of the hypotenuse. So the distance squared plus 64 = 100. Subtracting sixty-four leaves thirty-six, and its square root is six. The chord therefore lies six centimetres from the centre. Notice that the right-angle mark alone would not justify halving the chord. The perpendicular must also pass through the centre. An unrelated perpendicular could meet the chord away from its midpoint.

3:09Always check both the angle and the position of the line before using this property.

Equal chords and distances

3:14The distance from a point to a line means the perpendicular distance. For a chord, we therefore measure straight from the centre to the chord at ninety degrees. A sloping segment would be longer and would not represent this distance. This definition is essential when comparing how far chords lie from the centre. Suppose two chords in the same circle have equal lengths. Draw a perpendicular from the centre to each chord.

3:37Each perpendicular bisects its chord, so the corresponding half-chords are equal. Joining the centre to one endpoint of each chord creates two right-angled triangles. These triangles have equal hypotenuses because both are radii of the same circle. They also have equal half-chords, so the triangles match. Their remaining sides, the perpendicular distances, must therefore be equal. That gives our next property: equal chords are equidistant from the centre. Equidistant simply means at the same distance.

4:06The chords do not need to be parallel for this property to work. One could be horizontal and the other tilted, provided their lengths are equal. Turning a chord around the centre changes its direction, but not its length or perpendicular distance. For our second example, two chords in the same circle have equal lengths. Their perpendicular distances from the centre are 2x + 1 centimetres and x + 5 centimetres.

4:29Equal chords are equidistant from the centre. We therefore write 2x + 1 = x + 5. Subtracting x from both sides gives x + 1 = 5. Then subtracting one gives x = 4. Substitution into the first distance gives two multiplied by four plus one, which equals nine centimetres. Checking the other distance gives 4 + 5, also nine centimetres. The phrase in the same circle is important because the proof depends on equal radii.

4:57The relationship also works between circles of equal radius. Equal chords in circles of different sizes need not have equal distances from their centres. In an examination, identify the relevant circle before forming your equation.

Equal tangent lengths

5:10Now consider two tangents drawn from one external point T. They touch the circle at P and Q. We need one previously learned property: a tangent is perpendicular to the radius at its point of contact. Joining O to both contact points therefore creates two right-angled triangles. Triangle OPT and triangle OQT share the hypotenuse from O to T. Their radii are also equal, so the triangles match. The corresponding tangent segments must have equal lengths.

5:36This explains the property that tangents from an external point are equal in length. The equal lengths run from the shared external point to the points of contact. We are not comparing entire tangent lines, because lines extend indefinitely. Nor are we comparing tangents drawn from unrelated external points. The shared starting point is what gives the two triangles their common hypotenuse. In our third example, the tangent lengths are 3x + 2 centimetres and 5x − 6 centimetres.

6:04We must calculate x and the combined length of the tangent segments. Tangents from an external point are equal in length. Therefore, 3x + 2 = 5x − 6. Subtracting 3x from both sides gives 2 = 2x − 6. Adding six gives 8 = 2x. Dividing by two gives x = 4. We now substitute this value into both original lengths. The upper tangent measures three multiplied by four plus two, giving fourteen centimetres.

6:33For the lower tangent, five multiplied by four minus six also gives fourteen centimetres. Their combined length is 14 + 14, which equals twenty-eight centimetres. Checking both lengths confirms that our answer satisfies the geometrical property. Our fourth example joins this property to Pythagoras' theorem. A circle has centre O and a radius of five centimetres. An external point T lies thirteen centimetres from O, and its tangents touch the circle at P and Q.

6:59We must calculate the length from T to P and the area of the quadrilateral OPTQ. A tangent is perpendicular to the radius at the point of contact, so triangle OPT has a right angle at P. Its hypotenuse is the line from O to T. Subtracting twenty-five from one hundred and sixty-nine gives one hundred and forty-four. The square root of this is twelve, so the tangent from T to P is twelve centimetres long.

7:24Tangents from an external point are equal in length, so the tangent from T to Q is also twelve centimetres. The quadrilateral is a kite made of two matching right-angled triangles. Each triangle has an area of one half of 5 × 12, which is thirty square centimetres. The kite therefore has an area of sixty square centimetres. You must be able to carry out all of these examples without a calculator on Paper two, because each square root is exact.

7:49The same reasoning appears on calculator Paper four, where a square root is often not exact. Keep intermediate values unrounded there. Give final answers to three significant figures unless the question asks for something else. In examination questions, the command calculate requires a numerical answer supported by working. If asked to give a reason, state the specific geometrical property used. Saying only circle theorem does not identify the relationship. Write that equal chords are equidistant from the centre, or that tangents from an external point are equal in length.

8:19Avoid trusting a diagram simply because two lengths look equal. Look for stated information, equality marks, a centre label, or a tangent description. A perpendicular bisector needs both a right angle and equal halves. A centre-to-chord distance must be perpendicular, and equal tangent segments must share an external point. All three properties come from the same underlying idea: equal radii produce matching triangles and symmetry. A chord’s perpendicular bisector passes through the centre.

8:48Equal chords lie at equal perpendicular distances from that centre. Tangent segments from one external point have equal lengths. These facts stay reliable even when a diagram is not drawn to scale.