A tangent to a circle at a given point is the straight line that touches the circle exactly there and is perpendicular to the radius at that point. To find its equation: calculate the gradient of the radius, take its negative reciprocal for the tangent gradient, then form the straight-line equation through the point of contact.
What is the key principle behind the method?
The key fact is: a tangent to a circle is perpendicular to the radius at the point of contact. This is a circle theorem (the tangent-radius theorem).
So if you know:
- The centre of the circle and the point on the circle → you can find the gradient of the radius.
- The perpendicular gradient rule → you can find the gradient of the tangent.
- The coordinates of the point of contact → you can find the equation of the tangent.
How do you find the equation of a tangent to a circle at the origin?
For a circle with equation x² + y² = r² (centre at the origin), the tangent at point (x₁, y₁) is:
x₁x + y₁y = r²
This is a standard result worth memorising for Higher GCSE.
Worked example 1: Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4).
Check (3, 4) is on the circle: 3² + 4² = 9 + 16 = 25. ✓
Using the formula: tangent equation is 3x + 4y = 25.
Alternatively, using first principles:
- Radius from (0, 0) to (3, 4): gradient = (4 − 0)/(3 − 0) = 4/3
- Tangent gradient (perpendicular): −3/4
- Equation through (3, 4) with gradient −3/4: y − 4 = −3/4(x − 3) → 4y − 16 = −3x + 9 → 3x + 4y = 25 ✓
How do you find the tangent to a circle not centred at the origin?
For a circle (x − a)² + (y − b)² = r² with centre (a, b), use the first-principles approach.
Step 1: Find the gradient of the radius from the centre (a, b) to the point (x₁, y₁).
Gradient of radius = (y₁ − b) / (x₁ − a)
Step 2: The tangent is perpendicular to the radius, so:
Gradient of tangent = −(x₁ − a) / (y₁ − b)
Step 3: Use the point (x₁, y₁) and the tangent gradient in y − y₁ = m(x − x₁).
Step 4: Rearrange to the required form (usually y = mx + c or ax + by = c).
Worked examples
Worked example 2: Find the equation of the tangent to (x − 2)² + (y + 1)² = 50 at the point (9, 6).
First, check the point is on the circle: (9 − 2)² + (6 + 1)² = 49 + 49 = 98. This does not equal 50 — the point is not on the circle as given. Let's correct the example: use radius² = 98, so the circle is (x − 2)² + (y + 1)² = 98.
Centre (2, −1), point (9, 6).
Gradient of radius = (6 − (−1)) / (9 − 2) = 7/7 = 1
Gradient of tangent = −1 (negative reciprocal of 1)
Equation of tangent through (9, 6) with gradient −1:
y − 6 = −1(x − 9) → y − 6 = −x + 9 → y = −x + 15 (or x + y = 15)
Worked example 3: The circle has equation x² + y² − 4x + 6y = 12. Find the equation of the tangent at the point (5, 1).
First complete the square to find the centre: (x² − 4x + 4) + (y² + 6y + 9) = 12 + 4 + 9 (x − 2)² + (y + 3)² = 25
Centre: (2, −3). Radius: 5.
Check (5, 1): (5 − 2)² + (1 + 3)² = 9 + 16 = 25. ✓
Gradient of radius from (2, −3) to (5, 1) = (1 − (−3))/(5 − 2) = 4/3
Gradient of tangent = −3/4
Equation through (5, 1): y − 1 = −3/4(x − 5) → 4y − 4 = −3x + 15 → 3x + 4y = 19
What are the steps in a clear method summary?
| Step | Action |
|---|---|
| 1 | Identify the centre (a, b) and the point of tangency (x₁, y₁) |
| 2 | Confirm (x₁, y₁) lies on the circle (check in the equation) |
| 3 | Gradient of radius = (y₁ − b)/(x₁ − a) |
| 4 | Gradient of tangent = −(x₁ − a)/(y₁ − b) |
| 5 | Equation of tangent: y − y₁ = m(x − x₁); rearrange |
Frequently asked questions
Do I need to check that the point is on the circle?
Yes — always verify. If the point is not on the circle, the question has an error or you have misread it. Substituting the given coordinates into the circle equation should give the right-hand side exactly. If it does not, recheck your reading of the question before proceeding.
What if the radius is vertical (gradient undefined)?
If x₁ = a (the point of tangency is directly above or below the centre), the radius is vertical. A line perpendicular to a vertical line is horizontal, so the tangent equation is simply y = y₁.
How does this relate to the circle theorem about tangents?
The tangent-radius perpendicularity is one of the circle theorems required at GCSE. Here it is applied algebraically rather than geometrically, but it is the same underlying fact. In a geometry question you would state "tangent is perpendicular to the radius" as your reason; in a coordinate geometry question you apply it by taking the negative reciprocal of the radius gradient.
Is this Higher only?
Yes. Finding the equation of a tangent to a circle using coordinate methods is a Higher tier topic at GCSE. It combines the equation of a circle (also Higher) with perpendicular lines and straight-line equations, all of which are individually testable topics at Higher tier.
For Socratic GCSE Higher coordinate geometry practice with Professor Pi, see aitutors.me.