Read the core explanations from Tangents to Circles before testing yourself from memory.
1. Why Tangents to Circles Matter
Picture a ball rolling along a curved track. At any instant, the direction it travels is along the tangent to the curve at that point. In GCSE maths, you need to find the equation of that tangent line when the curve is a circle — and there is a clean, reliable method to do it every time. In addition, understanding the concept of tangency is essential because it not only provides key geometric insights but also integrates algebraic methods with spatial reasoning. Recognising how the radius and tangent are related forms the basis for many geometric problems, and a deeper grasp of these ideas strengthens problem-solving skills for both exams and practical applications.
The key insight is geometric: a tangent to a circle is always perpendicular to the radius at the point of contact. That single fact drives the entire method. Once you see why, the algebra follows naturally.
2. The Core Method: Three Steps to the Tangent Equation
Every question of this type follows the same three-step process. Learn the steps in order and you have a reliable method for any circle, any point.
- Step 1 — Find the gradient of the radius from the centre to the given point on the circle.
- Step 2 — Use the perpendicular gradient rule to find the gradient of the tangent.
- Step 3 — Use $y - y_1 = m(x - x_1)$ with the tangent gradient and the given point to write the equation.
Before working through a full example, make sure the perpendicular gradient rule is solid. If the radius has gradient $\frac{3}{4}$, the tangent has gradient $-\frac{4}{3}$. You flip the fraction and change the sign — that is all.
3. Worked Example — Circle Centred at the Origin
The most common exam setup gives you a circle centred at the origin $O(0, 0)$ and a point on the circle. Here is a full walkthrough so you can see every step in action.
Worked example: Finding the Equation of a Tangent at a Given Point
A circle is centred at the origin $O(0, 0)$ and passes through the point $P(3, 4)$. Find the equation of the tangent to the circle at $P$. Give your answer in the form $y = mx + c$.
- $m_{\text{radius}} = \frac{4 - 0}{3 - 0} = \frac{4}{3}$ — Use $m = \frac{y_2 - y_1}{x_2 - x_1}$ with $O(0,0)$ and $P(3,4)$. [M1 — correct gradient method]
- $m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4}$ — The tangent is perpendicular to the radius, so use the negative reciprocal. [M1 — perpendicular gradient]
- $y - 4 = -\frac{3}{4}(x - 3)$ — Substitute $m = -\frac{3}{4}$ and the point $(3, 4)$ into $y - y_1 = m(x - x_1)$. [M1 — correct substitution]
- $y - 4 = -\frac{3}{4}x + \frac{9}{4}$
$y = -\frac{3}{4}x + \frac{9}{4} + 4$
$y = -\frac{3}{4}x + \frac{9}{4} + \frac{16}{4}$
$y = -\frac{3}{4}x + \frac{25}{4}$ — Add 4 (written as $\frac{16}{4}$) to both sides. Keep fractions exact — do not convert to decimals mid-working. [A1 — correct final equation]
- $y = -\frac{3}{4}(3) + \frac{25}{4} = -\frac{9}{4} + \frac{25}{4} = \frac{16}{4} = 4$ ✓ — Substituting $x = 3$ gives $y = 4$, which matches point $P(3,4)$. If this check fails, an arithmetic error has been made somewhere above.
Answer: $y = -\frac{3}{4}x + \frac{25}{4}$
Notice how the sanity check in Step 5 costs almost no time but catches arithmetic slips before they cost marks. Build this habit into every tangent question.
4. Worked Example — Circle with a Non-Origin Centre
GCSE Higher exam questions sometimes shift the centre away from the origin. The method is identical — you need to use the actual centre coordinates when finding the radius gradient in Step 1.
Worked example: Tangent to a Circle with Centre Not at the Origin
A circle has centre $C(2, -1)$ and passes through the point $Q(6, 2)$. Find the equation of the tangent to the circle at $Q$.
- $m_{\text{radius}} = \frac{2 - (-1)}{6 - 2} = \frac{3}{4}$ — Use $m = \frac{y_2 - y_1}{x_2 - x_1}$ with $C(2,-1)$ and $Q(6,2)$. Note: $2 - (-1) = 3$, not $1$. [M1]
- $m_{\text{tangent}} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}$ — Negative reciprocal of $\frac{3}{4}$ is $-\frac{4}{3}$. [M1]
- $y - 2 = -\frac{4}{3}(x - 6)$ — Substitute $m = -\frac{4}{3}$ and point $Q(6, 2)$ into $y - y_1 = m(x - x_1)$. [M1]
- $y - 2 = -\frac{4}{3}x + 8$
$y = -\frac{4}{3}x + 8 + 2$
$y = -\frac{4}{3}x + 10$ — $-\frac{4}{3} \times (-6) = +8$. Then add 2 to both sides. [A1]
- $y = -\frac{4}{3}(6) + 10 = -8 + 10 = 2$ ✓ — Substituting $x = 6$ gives $y = 2$, confirming $Q(6, 2)$ lies on the tangent line.
Answer: $y = -\frac{4}{3}x + 10$
5. Guided Practice
Now it is your turn. Work through these problems using the three-step method. Cover the hints and try each one before checking.
6. Exam Technique and Common Mistakes
This topic rewards students who follow the method carefully. Most marks are lost not through misunderstanding the concept, but through small errors in the algebra. Here is where those errors tend to appear — and how to avoid them. Moreover, practising the full process thoroughly builds confidence and highlights common pitfalls, such as miscalculating the gradient or failing to verify that the tangent line actually touches the circle at the correct point. By elaborating on each step and understanding the cause of frequent mistakes, students are better prepared for both timed exams and challenging problem sets.
7. Exam-Style Question
This question is written in Higher-tier exam style. Work through it independently, then check your answer against the marking points below.
Every tangent-to-a-circle question on GCSE Higher reduces to the same three steps: radius gradient, perpendicular gradient, line equation. Practise the method until it is automatic, and the algebra becomes the only variable.