Equation of a Circle & Tangent to a Circle at a Given Point

SPEC A16 • ALGEBRA

Equation of a Circle & Tangent at a Point

Higher tier only. Recognise x² + y² = r² as a circle centred at the origin, and find the equation of the tangent to that circle at a given point using perpendicular gradients.

Two Skills, Very Different Difficulty

This topic splits neatly into two question types:

Part (a) — Usually 1–2 marks

Write the equation of the circle. Straightforward if you know the form, but there's a catch: the question rarely gives you the radius directly. It might give you a point on the circle (Jun 2024 P1H, Nov 2022 P3H) or the circumference (Nov 2024 P1H) — you extract the radius first, then square it.

Part (b) — Usually 3–5 marks

Find the equation of the tangent at a given point — and sometimes a follow-up like "where does the tangent cross the x-axis?" (Nov 2022 P1H). Combines circle geometry, gradient calculation, and perpendicular line equations. A grade 9 discriminator.

How often does this come up?

In the four most recent AQA sessions (Jun 2022 to Nov 2024), a circle-equation question has appeared in every single series. It's one of the highest-yield Higher-tier topics — learn it well and it is almost guaranteed marks.

Key Terms

Radius

A line from the centre of the circle to any point on the circle. Its length is r. Every point on the circle is distance r from the centre.

Tangent to a circle

A straight line that touches the circle at exactly one point. The tangent is always perpendicular to the radius drawn to the point of contact — this is the key property for finding its equation.

Perpendicular gradients

Two lines are perpendicular when the product of their gradients is −1. If one line has gradient m, the perpendicular gradient is −1/ m — the negative reciprocal. (See lesson 2.12 for full treatment.)

Where x² + y² = r² Comes From

This is Pythagoras' theorem in disguise. Take any point (x, y) on a circle centred at the origin with radius r. Drop a perpendicular to the x-axis:

The radius r (red) is the hypotenuse of a right triangle with horizontal leg x and vertical leg y. Pythagoras gives x² + y² = r² — true for every point on the circle.

Important — centre at origin only

This form only works when the centre is (0, 0). Circles with centre (a, b) have equation (x − a)² + (y − b)² = r² — but that is A-level, not GCSE. At GCSE, every circle has centre at the origin.

Reading the Equation: r² vs r

The number on the right-hand side of the equation is r ², not r itself. Take the square root to find the radius.

Equationr ² =Radius r =

x ² + y ² = 25255

x ² + y ² = 49497

x ² + y ² = 32432418

x ² + y ² = 202020 = 2 5

The question rarely gives you the radius

Recent AQA papers have given students a point on the circle or the circumference — never the radius directly. You have to extract the radius first. Examples:

Passes through (0, 6): the distance from origin is the radius → r = 6 → x² + y² = 36.

Circumference 36π: use 2πr = 36π → r = 18 → x² + y² = 324. (Watch out: if you accidentally use π r² = 36π — the area formula — you get r = 6 instead of 18. This was the most common error on Nov 2024 P1H Q17.)

Diameter 12: halve it → r = 6 → x² + y² = 36 (not 144).

Worked Example 1: Write the Equation

Ex1

Higher — 1 mark (Jun 2024 P1H Q18 pattern)

A circle has centre O and passes through (0, 6). Write down the equation of the circle.

MMethod. Circle centred at origin has form x² + y² = r². The given point is (0, 6) — on the y-axis, 6 units from the origin. So the radius is 6.

CCalculate.r² = 6² = 36.

EEvaluate. Equation: x ² + y ² = 36.

Variant: circumference given (2 marks)

If the question gave the circumference instead (e.g. Nov 2024 P1H Q17: circumference 36π), the first mark is for extracting the radius: 2πr = 36π → r = 18. Then the second mark: x² + y² = 18² = 324. The "method" step becomes the first M1 in a 2-mark question.

Tangent Property: Perpendicular to Radius

At any point on a circle, the tangent is perpendicular to the radius drawn to that point. This is the key geometric fact for the whole tangent question.

Circle x² + y² = 25, radius 5. At point P(3, 4), the tangent (light blue) meets the radius (dark blue) at a right angle — marked with the small square.

Gradient of the Radius

For a circle centred at the origin, the radius to point (a, b) is a straight line from (0, 0) to (a, b). Using rise over run:

gradient of radius = (b − 0) / (a − 0) = b / a

Example — point (3, 4)

Gradient of the radius from origin to (3, 4) = 4/3.

(Rise of 4 units, run of 3 units.)

Memory hook

For a circle at origin, the radius gradient is just the point's coordinates as a fraction — y-coordinate over x-coordinate. Simple as that.

Tangent Gradient: Negative Reciprocal

Because the tangent is perpendicular to the radius, its gradient is the negative reciprocal of the radius gradient:

gradient of tangent = − a / b

Example — point (3, 4)

Radius gradient = 4/3. Tangent gradient = −3/4.

(Flip the fraction, and change the sign.)

Check: (4/3) × (−3/4) = −12/12 = −1 ✓ (perpendicular condition satisfied)

Flip AND negate — not just one

The negative reciprocal requires both steps: flip the fraction (4/3 → 3/4) andchange the sign (3/4 → −3/4). Doing only one of these is the single most common mistake on tangent questions.

MCE Chain for Tangent Questions

Every tangent-to-a-circle question follows the same four-stage process. The mark scheme typically awards a mark for each stage — state each one clearly.

MMethod. (1) Find gradient of radius from origin to the given point: gradient = b/a. (2) Find tangent gradient as negative reciprocal: −a/b. (3) Use point-slope form y − b = m(x − a) with the tangent gradient and given point.

CCalculate. Substitute the numbers and expand carefully. Rearrange to a clean form — usually ax + by = r² or y = mx + c as the question demands.

EEvaluate.Verify by substituting the given point back into your final tangent equation — it must satisfy it. Check the gradient matches what you calculated. State the tangent equation clearly.

Worked Example 2: Setup

Ex2

Higher — 4 marks

Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4).

M — Method

First, verify the given point is actually on the circle: 3² + 4² = 9 + 16 = 25 ✓. Good.

The plan:

Find gradient of radius from (0, 0) to (3, 4): b/a = 4/3.

Tangent gradient = negative reciprocal of 4/3 = −3/4.

Use point-slope form y − 4 = (−3/4)(x − 3) with the tangent gradient and point (3, 4).

Worked Example 2: Calculate

C — Calculate

Start: y − 4 = (−3/4)(x − 3)

Multiply both sides by 4 (to clear the fraction):

4(y − 4) = −3(x − 3)

Expand brackets:

4y − 16 = −3x + 9

Rearrange to ax + by = c form:

3x + 4y = 9 + 16

3x + 4y = 25

Neat trick

For a circle centred at origin, the tangent at (a, b) always has equation ax + by = r ². So for (3, 4) on x²+y²=25, the tangent is 3x + 4y = 25 — you can spot this by inspection. But in an exam, always show the full working: method marks come from the process, not the answer alone.

Worked Example 2: Evaluate & Verify

E — Evaluate

Verification step: substitute (3, 4) into the tangent equation 3x + 4y = 25: LHS = 3(3) + 4(4)

= 9 + 16

= 25 = RHS ✓ The point (3, 4) lies on the tangent line as required.

Final Answer

3x + 4y = 25

(or equivalently, y = −3/4 x + 25/4)

Mark breakdown

M1 gradient of radius (4/3). M1 perpendicular gradient (−3/4). M1 using point-slope with tangent gradient and point. A1 correct final equation in any valid form.

Weak vs Strong: The Parallel Trap

Weak student

Uses the radius gradient (4/3) as the tangent gradient. Writes tangent as y = (4/3)x + c.

This makes the tangent PARALLEL to the radius — which is geometrically impossible for a tangent.

Strong student

Takes the negative reciprocal: flips 4/3 to 3/4, then negates to −3/4. Uses that as the tangent gradient.

Creates a proper perpendicular line — passes through the point, touches circle once only.

Quick self-check

Multiply your radius and tangent gradients together. The answer must be −1. If it isn't, you've either forgotten to flip or forgotten to negate. Fix before moving on — this single check catches almost every perpendicular-gradient error.

Examiner Alerts

Nov 2024 P1H Q17(a) — circumference vs area

The question gave circumference 36π and asked for the circle's equation. The correct route is 2πr = 36π → r = 18 → x² + y² = 324. The most common error was using the area formula πr² = 36π, which gives r = 6 — so students wrote x² + y² = 36 instead of 324. Decide which formula applies before solving.

Forgetting to square the radius

The other common single-mark killer: writing x² + y² = r instead of r². For a circle of radius 5, the equation is x² + y² = 25, notx² + y² = 5. Mark scheme explicitly gives M1A0 for this.

Sign errors on the reciprocal

The tangent gradient flips AND negates the radius gradient. Writing just −(4/3) or just (3/4) without the minus sign scores zero on the "perpendicular gradient" method mark. Make the sign change explicit in your working.

Skipping the verification

Your tangent equation must be satisfied by the given point. If 3x + 4y = 25 and the point is (3, 4), you should get 9 + 16 = 25 when you check. If the check fails, there is an arithmetic error — find it before moving on.

Practice Question 1

Write the equation of a circle.

(a) A circle has centre (0, 0) and radius 7. Write its equation. [1 mark]

(b) A circle has centre (0, 0) and diameter 12. Write its equation. [2 marks]

M

Form: x² + y² = r². (a) r = 7 is given directly. (b) Diameter is 12, so radius = 6 — this is where the mark is.

C

(a) r² = 7² = 49. (b) r² = 6² = 36.

E

(a) x ² + y ² = 49. (b) x ² + y ² = 36. The trap on (b) was the diameter — many students would write 144 by mistake.

Practice Question 2

Tangent to a circle. The point P(6, 8) lies on the circle x² + y² = 100. Find the equation of the tangent to the circle at P. [4 marks]

M

Verify P on circle: 6² + 8² = 36 + 64 = 100 ✓. Radius gradient = 8/6 = 4/3. Tangent gradient = negative reciprocal = −3/4.

C

y − 8 = (−3/4)(x − 6). Multiply by 4: 4y − 32 = −3x + 18. Rearrange: 3 x + 4 y = 50.

E

Verify: 3(6) + 4(8) = 18 + 32 = 50 ✓. Gradient check: −3/4 × 4/3 = −1 ✓. Tangent: 3 x + 4 y = 50.

Practice Question 3

Mixed — building to authentic Nov 2022 P1H Q22 style. The circle C has equation x² + y² = 169.

(a) State the radius of C. [1 mark]

(b) Show that the point (5, 12) lies on C. [1 mark]

(c) Find the equation of the tangent to C at the point (5, 12). Give your answer in the form ax + by = c. [3 marks]

(d) The tangent in (c) intersects the x-axis at point P. Work out the x-coordinate of P. [2 marks]

(a)

r² = 169, so r = 169 = 13.

(b)

Substitute (5, 12): 5² + 12² = 25 + 144 = 169 ✓. Therefore (5, 12) lies on C.

(c) M

Radius gradient = 12/5. Tangent gradient = −5/12 (flip and negate).

(c) C

y − 12 = (−5/12)(x − 5). Multiply by 12: 12y − 144 = −5x + 25. Rearrange: 5 x + 12 y = 169.

(c) E

Verify: 5(5) + 12(12) = 25 + 144 = 169 ✓. Tangent: 5 x + 12 y = 169.

(d)

Tangent meets x-axis where y = 0. Substitute: 5x + 12(0) = 169 → 5x = 169 → x = 169/5 = 33.8. So P has x-coordinate 33.8 (or 169/5).

Summary

Circle Equation (Centre at Origin)

x² + y² = r²

The number on the RHS is r ² — take the square root for the radius.

Radius is half the diameter — always check which one the question gives.

Only centre-at-origin is in the GCSE spec; other centres are A-level.

Tangent at Point (a, b)

1

Radius gradient from origin: b / a

2

Tangent gradient (negative reciprocal): − a / b — flip AND negate

3

Point-slope form: y − b = (− a / b )( x − a )

4

Rearrange to ax + by = r ² — a shortcut for checking

5

Verify the point satisfies the equation

Self-Check

Point on circle? (a² + b² = r²?)

Gradients multiply to −1?

Point satisfies final equation?

Avoid the Traps

Radius ≠ diameter

r² ≠ r

Flip AND negate — not one or the other

Top Exam Tip

The final tangent to x² + y² = r² at (a, b) is always ax + by = r ². You can spot-check your answer against this form — but always show the full working (radius gradient → negative reciprocal → point-slope → rearrange) to secure the method marks.