Tangents to Circles — Edexcel A-level Mathematics (Pure)

SPEC 3.2 • COORDINATE GEOMETRY

Chord and Tangent Properties

Using geometric properties to find equations of tangents and work with chords

Tangent Properties

Definition

A tangent to a circle is a straight line that touches the circle at exactly one point.

Key Property

A tangent to a circle is perpendicular to the radius at the point of contact.

If the gradient of the radius is

m

, then the gradient of the tangent is

- \frac{1}{m}

Chord Properties

Definition

A chord is a line segment that joins two points on the circumference of a circle.

Key Property

The perpendicular bisector of any chord passes through the centre of the circle.

This means:

The line from centre ⊥ to a chord bisects the chord

The line from centre to midpoint of chord is ⊥ to the chord

Finding Tangent at a Point

When given a point

P

on the circle, use the perpendicular relationship:

Method 1: Tangent at a Given Point

1

Find the centre of the circle (from equation)

2

Find the gradient of radius CP using

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

3

Find gradient of tangent =

- \frac{1}{m}

(negative reciprocal)

4

Use

y - y_{1} = m_{tangent} (x - x_{1})

with point

P

Remember

The tangent is perpendicular to the radius at that point. If the gradient of the radius is

m

, then the gradient of the tangent will be

- \frac{1}{m}

01Worked Example

The circle $C$ has equation $(x - 2)^{2} + (y - 6)^{2} = 100$ .

a) Verify that the point $P (10, 0)$ lies on $C$ .

b) Find an equation of the tangent to $C$ at the point $(10, 0)$ , giving your answer in the form $a x + b y + c = 0$ .

Part (a): Substitute

(x, y) = (10, 0)

into the equation:

•

1.

(10 - 2)^{2} + (0 - 6)^{2} = 8^{2} + (- 6)^{2}

•

2.

= 64 + 36 = 100

✓

Part (b): The centre is

(2, 6)

•

3. Gradient of radius =

\frac{6 - 0}{2 - 10} = \frac{6}{- 8} = - \frac{3}{4}

•

4. Gradient of tangent =

\frac{4}{3}

(negative reciprocal)

•

5.

y - 0 = \frac{4}{3} (x - 10)

•

6.

3 y = 4 x - 40

Answer: $4 x - 3 y - 40 = 0$

Tangent with Given Gradient

Sometimes you're given the gradient of the tangent and need to find the equation(s). There will be two parallel tangents with any given gradient.

Method 2: Given Gradient

1

Write tangent as

y = m x + c

2

Substitute into circle equation

3

Use discriminant = 0 (tangent touches once)

4

Solve for

c

Key Point

For a line to be tangent to a circle, the resulting quadratic must have exactly one solution, so

b^{2} - 4 a c = 0

Worked Example02

A circle $C$ has equation $(x - 5)^{2} + (y + 3)^{2} = 10$ . The line $l$ is a tangent to the circle and has gradient $- 3$ . Find two possible equations for $l$ , giving your answers in the form $y = m x + c$ .

•

1. Let tangent be

y = - 3 x + c

. Substitute into circle equation.

•

2.

(x - 5)^{2} + (- 3 x + c + 3)^{2} = 10

•

3.

x^{2} - 10 x + 25 + 9 x^{2} - 6 (c + 3) x + (c + 3)^{2} = 10

•

4.

10 x^{2} - (10 + 6 c + 18) x + 25 + (c + 3)^{2} - 10 = 0

•

5.

10 x^{2} - (6 c + 28) x + (c + 3)^{2} + 15 = 0

•

6. Discriminant = 0:

(6 c + 28)^{2} - 40 ((c + 3)^{2} + 15) = 0

•

7.

36 c^{2} + 336 c + 784 - 40 c^{2} - 240 c - 360 - 600 = 0

•

8.

- 4 c^{2} + 96 c - 176 = 0

, so

c^{2} - 24 c + 44 = 0

•

9.

(c - 2) (c - 22) = 0

, so

c = 2

or

c = 22

Answer: $y = - 3 x + 2$ and $y = - 3 x + 22$

Perpendicular Bisector of Chord

The perpendicular bisector of any chord passes through the centre. Use this to:

Applications

Find the centre of a circle given a chord

Show a line passes through the centre

Find circle equation from 3 points

Method

1

Find the midpoint of the chord

2

Find the gradient of the chord

3

Perpendicular gradient =

- \frac{1}{m}

4

Equation through midpoint with ⊥ gradient

03Worked Example

The points $P$ and $Q$ lie on a circle with centre $C$ . Point $P$ has coordinates $(- 7, - 1)$ and point $Q$ has coordinates $(3, - 5)$ . $M$ is the midpoint of $PQ$ . The line $l$ passes through $M$ and $C$ .

a) Find an equation for $l$ .

•

1. Midpoint

M = (\frac{- 7 + 3}{2}, \frac{- 1 + ( - 5 )}{2}) = (- 2, - 3)

•

2. Gradient of

PQ = \frac{- 5 - ( - 1 )}{3 - ( - 7 )} = \frac{- 4}{10} = - \frac{2}{5}

•

3. Line

l

is perpendicular to

PQ

(perpendicular bisector passes through centre)

•

4. Gradient of

l = \frac{5}{2}

(negative reciprocal)

•

5. Using point

M (- 2, - 3)

:

y + 3 = \frac{5}{2} (x + 2)

•

6.

y + 3 = \frac{5}{2} x + 5

Answer: $y = \frac{5}{2} x + 2$

04Worked Example

(Continuing from Example 3) Given that the $y$ -coordinate of $C$ is $- 8$ :

b) Show that the $x$ -coordinate of $C$ is $- 4$ .

c) Find an equation of the circle.

Part (b): The centre lies on line

l

:

y = \frac{5}{2} x + 2

•

1. Substitute

y = - 8

:

- 8 = \frac{5}{2} x + 2

•

2.

\frac{5}{2} x = - 10

, so

x = - 4

✓

Part (c): Centre is

C (- 4, - 8)

. Find radius using

CQ

:

•

3.

CQ = (3 - (- 4))^{2} + (- 5 - (- 8))^{2}

•

4.

= 49 + 9 = 58

Answer: $(x + 4)^{2} + (y + 8)^{2} = 58$

Circle Through Three Points

Given three points on a circle, you can find its equation using perpendicular bisectors.

Method

1

Form two chords from the three points

2

Find perpendicular bisector of each chord

3

Solve simultaneously to find centre

4

Find radius using distance to any point

5

Write the equation

Why This Works

The centre is equidistant from all points on the circle. The perpendicular bisector of a chord contains all points equidistant from the chord's endpoints.

Worked Example05

Find the equation of the circle passing through the points $A (1, 5)$ , $B (9, 1)$ and $D (10, 8)$ .

Perpendicular bisector of AB:

•

1. Midpoint of

A B = (\frac{1 + 9}{2}, \frac{5 + 1}{2}) = (5, 3)

•

2. Gradient of

A B = \frac{1 - 5}{9 - 1} = - \frac{1}{2}

, so perp gradient =

2

•

3. Line 1:

y - 3 = 2 (x - 5)

→

y = 2 x - 7

Perpendicular bisector of AD:

•

4. Midpoint of

A D = (\frac{1 + 10}{2}, \frac{5 + 8}{2}) = (5.5, 6.5)

•

5. Gradient of

A D = \frac{8 - 5}{10 - 1} = \frac{1}{3}

, so perp gradient =

- 3

•

6. Line 2:

y - 6.5 = - 3 (x - 5.5)

→

y = - 3 x + 23

Find centre (intersection):

•

7.

2 x - 7 = - 3 x + 23

→

5 x = 30

→

x = 6

,

y = 5

•

8. Radius:

C A = (6 - 1)^{2} + (5 - 5)^{2} = 5

Answer: $(x - 6)^{2} + (y - 5)^{2} = 25$

Tangents from External Point

From any external point, exactly two tangents can be drawn to a circle. These tangents have equal length.

Properties

RP = RQ

(equal tangent lengths)

Both tangents ⊥ to their radii

CPRQ

forms a kite (or square if

CP = PR

)

Exam Tip

If told CPRQ is a square, then

CP = PR

, meaning radius = tangent length, and angle

PCR = 45°

Worked Example06

The circle with centre $C$ has equation $(x - 2)^{2} + (y - 1)^{2} = 10$ . The tangents to the circle at points $P$ and $Q$ meet at the point $R$ with coordinates $(6, - 1)$ . Show that $CPRQ$ is a square.

•

1. Centre

C = (2, 1)

, radius

r = 10

•

2. Distance

CR = (6 - 2)^{2} + (- 1 - 1)^{2} = 16 + 4 = 20

•

3. For a square:

CR = r 2

(diagonal of square with side

r

)

•

4. Check:

r 2 = 10 \times 2 = 20

✓

•

5. Tangent length:

R P^{2} = C R^{2} - C P^{2} = 20 - 10 = 10

•

6. So

RP = 10 = r

✓

All sides equal $10$ , all angles 90°, so $CPRQ$ is a square ✓

07Worked Example

The circle $C$ has equation $x^{2} + 18 x + y^{2} - 2 y + 29 = 0$ .

a) Verify the point $P (- 7, - 6)$ lies on $C$ .

b) Find an equation for the tangent to $C$ at $P$ , in the form $y = m x + b$ .

Part (a):

(- 7)^{2} + 18 (- 7) + (- 6)^{2} - 2 (- 6) + 29

•

1.

= 49 - 126 + 36 + 12 + 29 = 0

✓

Part (b): Find centre from general form:

•

2.

2 g = 18 \Rightarrow g = 9

,

2 f = - 2 \Rightarrow f = - 1

•

3. Centre =

(- g, - f) = (- 9, 1)

•

4. Gradient of radius =

\frac{- 6 - 1}{- 7 - ( - 9 )} = \frac{- 7}{2}

•

5. Gradient of tangent =

\frac{2}{7}

•

6.

y + 6 = \frac{2}{7} (x + 7)

Answer: $y = \frac{2}{7} x - 4$

Practice

Question 1:

a) Find the equation of the circle.

b) Find the equation of the tangent to the circle at

P

.

a)

r = (4 - 1)^{2} + (6 - (- 2))^{2} = 9 + 64 = 73

Equation:

(x - 4)^{2} + (y - 6)^{2} = 73

b) Gradient of radius =

\frac{6 - ( - 2 )}{4 - 1} = \frac{8}{3}

Tangent gradient =

- \frac{3}{8}

y + 2 = - \frac{3}{8} (x - 1)

→ $3 x + 8 y + 13 = 0$

Practice

Question 2:

a) Find the equation of the perpendicular bisector of

A B

.

b) Show that the perpendicular bisector passes through the centre.

a) Midpoint =

(\frac{- 1 + 7}{2}, \frac{- 9 + ( - 5 )}{2}) = (3, - 7)

Gradient of

A B = \frac{- 5 - ( - 9 )}{7 - ( - 1 )} = \frac{4}{8} = \frac{1}{2}

Perpendicular gradient =

- 2

y + 7 = - 2 (x - 3)

→ $y = - 2 x - 1$

b) Centre is

(1, - 3)

. Check:

- 2 (1) - 1 = - 3

✓ Centre lies on line.

Practice

Question 3: A circle

C

has equation

(x + 5)^{2} + (y + 3)^{2} = 80

. The line

l

is a tangent to the circle and has gradient

2

. Find two possible equations for

l

.

Let tangent be

y = 2 x + c

. Substitute into circle:

(x + 5)^{2} + (2 x + c + 3)^{2} = 80

x^{2} + 10 x + 25 + 4 x^{2} + 4 (c + 3) x + (c + 3)^{2} = 80

5 x^{2} + (10 + 4 c + 12) x + 25 + (c + 3)^{2} - 80 = 0

5 x^{2} + (4 c + 22) x + (c + 3)^{2} - 55 = 0

Discriminant = 0:

(4 c + 22)^{2} - 20 ((c + 3)^{2} - 55) = 0

Solving:

c = 7

or

c = - 27

Answer: $y = 2 x + 7$ and $y = 2 x - 27$

Summary

Key Properties

PropertyApplication

Tangent ⊥ RadiusUse negative reciprocal for tangent gradient

Perp. bisector through centreFind centre from chord, or circle from 3 points

Discriminant = 0Find tangent equations with given gradient

Equal tangent lengthsFrom external point:

RP = RQ

Exam Tips

Always find/identify the centre first

Draw a diagram to visualise the geometry

For "two tangents" problems, expect two values of

c

For 3 points: use two perpendicular bisectors to find centre