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Equation of a Tangent

HigherHigher tier onlyAQAEdexcelOCR

Get to grips with equation of a tangent using these Higher GCSE Maths practice questions. The worksheet focuses on finding the equation of a tangent to a circle, and the accompanying mark scheme breaks down each solution clearly. Suitable for AQA, Edexcel and OCR. Download the questions and answers for free. A tangent is perpendicular to the radius at the point of contact.

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These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA, Edexcel and OCR specifications. Every worksheet comes with a full mark scheme.

This is a Higher tier only topic, so there's no Foundation paper — only the Higher worksheet and mark scheme below.

Higher

WorksheetHigher · PDFMark SchemeHigher · PDF

Topic overview

A tangent to a circle touches it at exactly one point and is perpendicular to the radius drawn to that point. That right angle is the key fact that makes the whole topic work.

So to find the equation of a tangent at a point on a circle centred at the origin, first find the gradient of the radius from the origin to that point. The tangent's gradient is then the negative reciprocal.

With the gradient known and a point on the line given, the equation follows in the usual way: substitute both into \(y = mx + c\) and solve for \(c\).

Revision notes

The radius and tangent are perpendicular

The radius from the origin to the point \((a, b)\) has gradient \(\frac{b}{a}\).

Because the tangent is perpendicular, its gradient is \(-\frac{a}{b}\), the negative reciprocal. This single relationship drives every question of this type.

Finding the equation

Use the tangent gradient with the given point to find the intercept.

At \((3, 4)\) on \(x^2 + y^2 = 25\): the radius gradient is \(\frac{4}{3}\), so the tangent gradient is \(-\frac{3}{4}\). Then \(4 = -\frac{3}{4}(3) + c\), giving \(c = \frac{25}{4}\).

Checking your answer

The point must satisfy your tangent equation, and the two gradients must multiply to \(-1\).

Both checks are quick, and together they catch nearly every error in this topic.

Key points

  • A tangent touches the circle at one point.
  • The tangent is perpendicular to the radius at that point.
  • Radius gradient at \((a,b)\) is \(\frac{b}{a}\).
  • Tangent gradient is \(-\frac{a}{b}\).
  • Substitute the point to find the intercept.
  • Check the gradients multiply to \(-1\).

Worked examples

Example 1

Find the gradient of the radius to the point \((4, 3)\) on a circle centred at the origin.

Working

\[\frac{3 - 0}{4 - 0}\]gradient from the origin to the point
\[= \tfrac{3}{4}\]state the radius gradient

Example 2

Find the gradient of the tangent at that point.

Working

\[\text{Negative reciprocal of } \tfrac{3}{4}\]the tangent is perpendicular to the radius
\[= -\tfrac{4}{3}\]flip and change the sign

Example 3

Find the equation of the tangent to \(x^2 + y^2 = 25\) at \((3, 4)\).

Working

\[\text{Radius gradient } \tfrac{4}{3} \text{, tangent } -\tfrac{3}{4}\]take the negative reciprocal
\[4 = -\tfrac{3}{4}(3) + c\]substitute the point
\[y = -\tfrac{3}{4}x + \tfrac{25}{4}\]solve for c and write the equation

Common mistakes

  • Using the radius gradient as the tangent gradient.

    The tangent is perpendicular, so take the negative reciprocal.

  • Flipping without changing the sign.

    Both steps are needed, or the gradients will not multiply to −1.

  • Substituting into the wrong equation.

    Use the tangent's gradient with the given point to find c.

  • Forgetting the point lies on both the circle and the tangent.

    It must satisfy both, which is a useful check.

Exam tips

  • Find the radius gradient first, then take its negative reciprocal.
  • Multiply the two gradients to check they give −1.
  • Substitute the point back into your tangent equation to verify.
  • Leave fractions as fractions rather than rounding.

Key terms

Tangent
A line touching a circle at exactly one point.
Radius
A line from the centre to a point on the circle.
Perpendicular
Meeting at a right angle.
Negative reciprocal
The result of flipping a fraction and reversing its sign.

Related topics

Written and reviewed against the current AQA, Edexcel and OCR specifications. Spotted an error? Let us know.