Chords and Tangent Geometry

Half-chord, radius and distance from centre make a right angle: chord length, or the line that gives a chord of a given length

In the exam for Edexcel International AS Maths · 6 real past-paper questions below

Chord lengths and the half-chord

You can already find where a line cuts a circle. AS questions add chord lengths, and they work backwards: they give the length and ask for the line.

One right angle does it. The line from the centre that meets a chord at right angles cuts the chord in half.

Formula

h^{2} + (\frac{L}{2})^{2} = r^{2}

$h$ — the distance from the centre to the chord, measured at right angles.
$\frac{L}{2}$ — half the chord length L.
$r$ — the radius. It is the longest side, from the centre to the end of the chord.

Worked example

1
Find the length of the chord cut from $(x - 2)^{2} + (y + 1)^{2} = 25$ by the line $y = 3$ .
2
$Centre (2, - 1), r = 5$ — Read them from the equation.
3
$h = 3 - (- 1) = 4$ — The chord is flat, so the distance is just the gap in y.
4
$(\frac{L}{2})^{2} = 25 - 16 = 9$ — Pythagoras: $r^{2} - h^{2}$ .
5
$\frac{L}{2} = 3, L = 6$ — Double the half. Check: the ends are (−1, 3) and (5, 3).

Now backwards. The length is given, and you want the line.

Worked example

1
A circle has centre $(2, - 3)$ and $r^{2} = 40$ . A horizontal chord above the centre has length $46$ . Find the equation of the chord.
2
$\frac{L}{2} = 26, (\frac{L}{2})^{2} = 24$ — Half of $46$ , then square it: $4 \times 6$ .
3
$h^{2} = 40 - 24 = 16, h = 4$ — Pythagoras again: $r^{2}$ minus the half-chord squared.
4
$y = - 3 + 4$ — Above the centre, so add h to the centre's y.
5
$y = 1$ — The other choice, $y = - 7$ , is the chord below the centre.

Common mistake

Using the whole chord in Pythagoras. The right angle is in the triangle with half the chord.

Rule

The mark scheme gives M1 for Pythagoras to find h, then M1 for centre ± h, then A1 for the exact answer. Read the condition, like “above the x-axis” or “p is negative”, to pick the sign.

Key idea

Centre, middle of the chord, end of the chord: a right-angled triangle with the radius as its longest side.

Find the centre from chords

The same right angle works the other way. The perpendicular bisector of any chord passes through the centre. It is the line through the chord's midpoint, at right angles to it.

Steps

1
Find the midpoint of the chord.
2
Find the chord's gradient, then flip it and change the sign.
3
Write the line through the midpoint with that gradient.
4
Do the same for a second chord. Where the two lines meet is the centre.

Worked example

1
$A (1, 7)$ , $B (9, 3)$ and $C (9, 11)$ lie on a circle. Find its centre and radius.
2
$M_{A B} = (5, 5), m_{A B} = - \frac{1}{2}$ — Average the coordinates. Gradient: down 4, across 8.
3
$y - 5 = 2 (x - 5) \Rightarrow y = 2 x - 5$ — Perpendicular gradient 2, through the midpoint.
4
$M_{A C} = (5, 9), m_{A C} = \frac{1}{2}$ — Up 4, across 8.
5
$y - 9 = - 2 (x - 5) \Rightarrow y = 19 - 2 x$ — Perpendicular gradient $- 2$ .
6
$2 x - 5 = 19 - 2 x \Rightarrow x = 6$ — Set equal: $4 x = 24$ .
7
$Centre (6, 7)$ — $y = 2 (6) - 5 = 7$ .
8
$r^{2} = (1 - 6)^{2} + (7 - 7)^{2} = 25, r = 5$ — Check with B: $3^{2} + (- 4)^{2} = 25$ .

Sometimes one chord is enough. If you are told the centre's x-coordinate, put it into the one perpendicular bisector to get its y.

Common mistake

Using the chord's own gradient for the bisector. The bisector is at right angles, so flip and change the sign.

Key idea

Archaeologists rebuild a whole plate from one broken piece this way. Draw two chords on the curved edge, then their perpendicular bisectors. They cross at the centre of the plate.

Key idea

Every chord's perpendicular bisector goes through the centre. Two of them pin it down.

Horizontal and vertical tangents

A horizontal or vertical tangent needs no algebra. It is exactly one radius from the centre.

Formula

y = b \pm r x = a \pm r

$b$ — the y-coordinate of the centre: the horizontal tangents are r above and r below it.
$a$ — the x-coordinate of the centre: the vertical tangents are r to the left and r to the right.
$r$ — the radius.

Memory trick

A circle in a box. Top and bottom walls: centre's y ± r. Left and right walls: centre's x ± r.

Worked example

1
The line $y = k$ is a tangent to $x^{2} + y^{2} - 4 x + 10 y + 11 = 0$ . Find the possible values of k.
2
$(x - 2)^{2} + (y + 5)^{2} = 18$ — Complete the square: $4 + 25 - 11 = 18$ .
3
$Centre (2, - 5), r = 32$ — $18 = 92$ .
4
$k = - 5 \pm 32$ — Horizontal tangent: centre's y ± r. Give both.

The vertical tangents would be $x = 2 \pm 32$ . You can also put $y = k$ into the circle and set the discriminant to 0. Both methods earn the marks; this one is quicker.

Common mistake

Adding r to the wrong coordinate. A horizontal line $y = k$ uses the centre's y.

Key idea

Flat or upright tangents: centre ± radius. Two answers each time.

From a tangent to the circle

You know the tangent is perpendicular to the radius at the touching point. AS questions use this to find that point when you have the tangent line and the centre.

Worked example

1
The line $y = 2 x + 2$ is a tangent to a circle with centre $X (5, 2)$ . Find the point N where it touches, and the equation of the circle.
2
$m_{X N} = - \frac{1}{2}$ — XN is a radius, so it is perpendicular to the tangent of gradient 2.
3
$y = - \frac{1}{2} x + \frac{9}{2}$ — Through the centre: $y - 2 = - \frac{1}{2} (x - 5)$ .
4
$2 x + 2 = - \frac{1}{2} x + \frac{9}{2}$ — N is on both lines, so set them equal.
5
$\frac{5}{2} x = \frac{5}{2} \Rightarrow N (1, 4)$ — $x = 1$ , then $y = 2 (1) + 2 = 4$ .
6
$r^{2} = (5 - 1)^{2} + (2 - 4)^{2} = 20$ — Distance from X to N.
7
$(x - 5)^{2} + (y - 2)^{2} = 20$ — The equation of the circle.

Another question type gives a point P on the circle and asks about the point opposite P, at the other end of the diameter. The centre is the midpoint, so step from P to the centre, then the same again.

Worked example

1
$P (- 1, 2)$ lies on $(x - 3)^{2} + (y + 1)^{2} = 25$ . Q is the point on the circle furthest from P. Find the tangent at Q in the form $a x + b y + c = 0$ .
2
$Q = (2 \times 3 - (- 1), 2 \times (- 1) - 2) = (7, - 4)$ — Opposite point: 2 × centre − P.
3
$m_{radius} = \frac{- 4 - ( - 1 )}{7 - 3} = - \frac{3}{4}$ — From the centre (3, −1) to Q.
4
$m_{tangent} = \frac{4}{3}$ — Flip and change the sign.
5
$y + 4 = \frac{4}{3} (x - 7)$ — Through Q, not the centre.
6
$4 x - 3 y - 40 = 0$ — Times 3: $3 y + 12 = 4 x - 28$ . Check: $28 + 12 - 40 = 0$ .

Common mistake

Writing the tangent through P instead of Q. Find Q first; most of the marks depend on it.

Rule

A “show that” part about the radius line needs every step: the perpendicular gradient, the point used, and the rearranging. If it then says “hence”, use that line to find the touching point.

Key idea

Tangent and radius meet at a right angle. The radius line through the centre finds the touching point, and the radius is the distance to it.

Which method? Read the question

Each part tells you which right angle to use.

You see	You do
A chord length, or a chord “of length …”	Pythagoras: $h^{2} + (\frac{L}{2})^{2} = r^{2}$ , then centre ± h
Three points on a circle, find the centre	Perpendicular bisectors of two chords; solve them together
A line through the midpoint of a chord and the centre	Negative reciprocal of the chord's gradient, through the midpoint
“y = k is a tangent” or “x = k is a tangent”	Centre's y ± r, or centre's x ± r
A tangent line and the centre are given	Radius line through the centre, perpendicular; meet the tangent
“Furthest from P” or “the other end of the diameter”	2 × centre − P; then the tangent there

Key idea

One fact runs through all of these: a line from the centre meets a chord or a tangent at a right angle.

Practise chords and tangent geometry now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on chords and tangent geometry

Each with a full worked solution.

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.