Finding the Equation of a Tangent to a Circle Revision Notes | GCSE AQA Higher Further Maths

1. Why Tangents to Circles Appear on Exam Papers

Picture a wheel rolling along a road. The road touches the wheel at exactly one point — and at that point, the road is perfectly perpendicular to the spoke pointing straight down. That relationship between a tangent and a radius is the entire engine behind this topic.

At GCSE, you will be given a circle with a known centre, a point on the circle, and asked to find the equation of the tangent at that point. The method is always the same four-step process, and once you see it clearly, it becomes one of the most reliable marks on the paper.

2. The Key Relationship: Tangent is Perpendicular to the Radius

This perpendicularity is the fact that makes the whole method work. You already know how to find the gradient of a line between two points, and you know that perpendicular lines have gradients that multiply to $-1$. Those two tools are all you need here.

3. The Four-Step Method

Every tangent-to-a-circle question follows the same four steps. Learn this sequence and you can handle any version the examiner sets.

• Step 1 — Find the gradient of the radius from the centre to the point of contact.
• Step 2 — Find the gradient of the tangent using the negative reciprocal.
• Step 3 — Substitute the tangent gradient and the point of contact into $y - y_1 = m(x - x_1)$.
• Step 4 — Simplify into the form $y = mx + c$ (or leave in the form the question requests).

Now watch this method in action on a full worked example. Try each step yourself before revealing it — that active effort is what makes the method stick.

Worked example: Finding the Equation of a Tangent to a Circle

A circle has centre $C(2, 1)$. The point $P(6, 4)$ lies on the circle. Find the equation of the tangent to the circle at $P$.

1. $m_{CP} = \frac{4 - 1}{6 - 2} = \frac{3}{4}$ — Use the gradient formula with $C(2,1)$ and $P(6,4)$. This earns a method mark [M1].
2. $m_{\text{tangent}} = -\frac{1}{m_{CP}} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}$ — The tangent is perpendicular to the radius, so flip the fraction and change the sign. [M1]
3. $y - 4 = -\frac{4}{3}(x - 6)$ — Use the point of contact $P(6, 4)$ and the tangent gradient $-\frac{4}{3}$. [M1]
4. $y - 4 = -\frac{4}{3}x + 8$ $y = -\frac{4}{3}x + 12$ — Add 4 to both sides: $-\frac{4}{3} \times (-6) = 8$, then $8 + 4 = 12$. Final answer [A1].

Answer: $y = -\frac{4}{3}x + 12$

4. A Second Worked Example — Circle Centred at the Origin

Many exam questions use a circle centred at the origin, $O(0, 0)$. The method is identical, but the gradient calculation is slightly quicker because one coordinate pair is $(0, 0)$.

Worked example: Tangent to a Circle Centred at the Origin

A circle has centre $O(0, 0)$ and passes through $Q(-5, 12)$. Find the equation of the tangent to the circle at $Q$. Give your answer in the form $y = mx + c$.

1. $m_{OQ} = \frac{12 - 0}{-5 - 0} = \frac{12}{-5} = -\frac{12}{5}$ — Gradient formula from $O(0,0)$ to $Q(-5, 12)$. [M1]
2. $m_{\text{tangent}} = -\frac{1}{-\frac{12}{5}} = \frac{5}{12}$ — Negative reciprocal of $-\frac{12}{5}$: flip to $\frac{5}{12}$ and the two negatives cancel to give a positive. [M1]
3. $y - 12 = \frac{5}{12}(x - (-5))$ $y - 12 = \frac{5}{12}(x + 5)$ — Use point $Q(-5, 12)$. Note the double negative: $x - (-5) = x + 5$. [M1]
4. $y - 12 = \frac{5}{12}x + \frac{25}{12}$ $y = \frac{5}{12}x + \frac{25}{12} + 12$ $y = \frac{5}{12}x + \frac{25}{12} + \frac{144}{12}$ $y = \frac{5}{12}x + \frac{169}{12}$ — Convert 12 to twelfths: $12 = \frac{144}{12}$. Then $\frac{25}{12} + \frac{144}{12} = \frac{169}{12}$. [A1]

Answer: $y = \frac{5}{12}x + \frac{169}{12}$

Notice how the double negative in Step 3 is a common slip point. When the $x$-coordinate of the point is negative, writing $x - (-5)$ and then simplifying to $x + 5$ is a step worth doing carefully and explicitly.

5. Guided Practice — Try These Yourself

The best way to consolidate this method is to work through questions with the steps in mind. Cover the solutions and attempt each one before checking.

6. Exam Technique and Common Mistakes

This topic rewards careful, methodical working. The marks are almost always awarded for the correct process, so showing each step clearly protects you even if arithmetic slips occur later.

One final check worth building into your routine: multiply the radius gradient by the tangent gradient. The product must equal $-1$. If it does not, you have made an error in Step 2 and should recheck before moving on.

7. A Harder Exam-Style Question

Exam papers sometimes give the circle equation in the form $(x - a)^2 + (y - b)^2 = r^2$ and ask you to read off the centre before applying the method. This is where reading the question carefully earns you the first mark for free.

Worked example: Tangent from a Circle Given in Standard Form

A circle has equation $(x - 3)^2 + (y + 2)^2 = 50$. The point $T(8, 3)$ lies on the circle. Find the equation of the tangent to the circle at $T$. Give your answer in the form $ax + by + c = 0$ where $a$, $b$, and $c$ are integers.

1. Centre $= (3, -2)$ — $(x-3)^2 + (y-(-2))^2 = 50$ gives centre $(3, -2)$. Note the sign: $+2$ inside the bracket means $y$-coordinate is $-2$. [B1]
2. $(8-3)^2 + (3-(-2))^2 = 5^2 + 5^2 = 25 + 25 = 50$ ✓ — Substituting $T(8,3)$ confirms it satisfies the equation. This is a useful sanity check.
3. $m_{\text{radius}} = \frac{3 - (-2)}{8 - 3} = \frac{5}{5} = 1$ — Gradient from centre $(3,-2)$ to $T(8,3)$. [M1]
4. $m_{\text{tangent}} = -\frac{1}{1} = -1$ — Negative reciprocal of $1$ is $-1$. [M1]
5. $y - 3 = -1(x - 8)$ $y - 3 = -x + 8$ $y = -x + 11$ — Point-gradient form with $T(8,3)$ and gradient $-1$. [M1]
6. $y = -x + 11$ $x + y - 11 = 0$ — Add $x$ and subtract $11$ from both sides. Here $a=1$, $b=1$, $c=-11$. [A1]

Answer: $x + y - 11 = 0$

The specification requirement — finding the equation of a tangent at a point on a circle — is fully covered by this four-step method. Whether the circle is centred at the origin, given by its centre coordinates, or presented in the form $(x-a)^2 + (y-b)^2 = r^2$, the process is always: radius gradient, negative reciprocal, point-gradient form, simplify.