A tangent touches a circle at one point, and it is perpendicular to the radius at that point. So to find a tangent, find the gradient of the radius, turn it into the perpendicular gradient, and use it with the point of contact.
This lesson belongs to circle coordinate methods and uses the perpendicular-gradient rule from straight lines and linearisation. Tangent questions appear alone and as part of longer circle problems.
What is the rule behind the method?
Two lines with gradients m₁ and m₂ are perpendicular when m₁ × m₂ = −1. So if the radius has gradient m, the tangent has gradient −1/m.
In words: flip the fraction and change the sign. A radius gradient of 4/3 gives a tangent gradient of −3/4.
How do you do it, step by step?
- Find the centre of the circle, completing the square if needed, and confirm the point really is on the circle.
- Find the gradient of the radius from the centre to the point.
- Flip and negate to get the tangent gradient.
- Use y − y₁ = m(x − x₁) with the point of contact and the tangent gradient.
- Tidy the equation and check that the point lies on it.
Worked example
Find the equation of the tangent to (x − 3)² + (y + 2)² = 25 at the point P(6, 2).
Step 1, centre and check: the centre is (3, −2). At P: (6 − 3)² + (2 + 2)² = 9 + 16 = 25, so P is on the circle. ✓
Step 2, radius gradient: from (3, −2) to (6, 2) the gradient is (2 − (−2)) ÷ (6 − 3) = 4/3.
Step 3, tangent gradient: −3/4.
Step 4, line through P: y − 2 = −3/4 (x − 6). Multiply by 4: 4y − 8 = −3x + 18.
Step 5, tidy: 3x + 4y = 26.
Check: P gives 18 + 8 = 26 ✓. The distance from the centre (3, −2) to the line is |9 − 8 − 26| ÷ 5 = 25 ÷ 5 = 5, which equals the radius. ✓
The mistake to watch for
The usual slip is to stop after the radius gradient and use it for the tangent.
Mistaken answer: y − 2 = 4/3 (x − 6), so 4x − 3y = 18
The student found the radius gradient 4/3 and used it directly. That line is the radius, not the tangent.
The correction is to add one extra line to every solution: “tangent gradient = −1 ÷ radius gradient”. Flipping without negating, giving 3/4 or −4/3, is the other common slip, so multiply the two gradients and confirm the product is −1.
Check yourself
1. Find the tangent to x² + y² = 13 at (2, 3).
Show answer
Radius gradient 3/2, tangent gradient −2/3. Then y − 3 = −2/3 (x − 2), so 3y − 9 = −2x + 4.
2x + 3y = 13. Check: 4 + 9 = 13. ✓
2. Find the tangent to (x − 1)² + (y − 2)² = 10 at (4, 3).
Show answer
Point check: 9 + 1 = 10 ✓. Radius gradient (3 − 2) ÷ (4 − 1) = 1/3, so tangent gradient −3.
y − 3 = −3(x − 4), so 3x + y = 15. Check: 12 + 3 = 15. ✓
3. Find the tangent to (x + 1)² + (y − 4)² = 9 at (2, 4).
Show answer
Point check: 9 + 0 = 9 ✓. The centre (−1, 4) and the point (2, 4) have the same y-coordinate, so the radius is horizontal. The tangent is vertical: x = 2.
Where this leads next
With tangents secure, move on to solving a parameter condition for tangency, then test the topic with the circle practice set. The non-calculator working trainer and the quadratic structure explorer help with the arithmetic and discriminant checks that follow.
If you understand why the tangent is perpendicular but still make sign errors under time pressure, a teacher can work on your routine in online one-to-one Additional Mathematics tuition.