Equation of a Tangent to a Circle

OCR

GCSE

The determination of the equation of a tangent to a circle relies fundamentally on the geometric property that the tangent is perpendicular to the radius at the point of contact. Candidates must calculate the gradient of the radius connecting the circle's centre $(a, b)$ to the point on the circumference, subsequently applying the negative reciprocal rule to establish the tangent's gradient. The final linear equation is constructed using the point-gradient form $y - y_1 = m(x - x_1)$. Advanced analysis involves using the discriminant of the combined linear and quadratic equations to prove tangency or identify unknown constants.

Objectives

Exam Tips

Pitfalls

Key Terms

Mark Points

What You Need to Demonstrate

Key skills and knowledge for this topic

Award M1 for correctly calculating the gradient of the radius using the center coordinates and the point of contact
Award M1 for applying the perpendicular gradient rule ($m_1 \times m_2 = -1$) to find the gradient of the tangent
Award M1 for substituting the point of contact and the tangent gradient into $y - y_1 = m(x - x_1)$ or $y = mx + c$
Award A1 for the correct final equation rearranged into the specified form (e.g., integer coefficients)

Example Examiner Feedback

Real feedback patterns examiners use when marking

"You correctly found the gradient of the radius, but you forgot to use the negative reciprocal for the tangent"
"Check your final rearrangement; the question asked for integer coefficients, but you left fractions in your answer"
"Good use of $y - y_1 = m(x - x_1)$; ensure you are substituting the point on the circumference, not the center of the circle"
"Your method is correct, but verify your arithmetic when calculating the gradient—check the signs of your coordinates"

Marking Points

Key points examiners look for in your answers

Award M1 for correctly calculating the gradient of the radius using the center coordinates and the point of contact
Award M1 for applying the perpendicular gradient rule ($m_1 \times m_2 = -1$) to find the gradient of the tangent
Award M1 for substituting the point of contact and the tangent gradient into $y - y_1 = m(x - x_1)$ or $y = mx + c$
Award A1 for the correct final equation rearranged into the specified form (e.g., integer coefficients)

Examiner Tips

Expert advice for maximising your marks

💡Always verify if the question requires the final answer in the form $ax + by + c = 0$ with integers; leaving fractions will lose the final mark
💡Draw a quick sketch of the circle and tangent to visually check if your calculated gradient should be positive or negative
💡When the center is not the origin, explicitly write down the coordinates of the center $(a,b)$ and the point $P$ before calculating the gradient to avoid substitution errors

Common Mistakes

Pitfalls to avoid in your exam answers

Using the gradient of the radius directly in the line equation rather than the negative reciprocal
Incorrectly calculating the gradient of the radius by swapping $x$ and $y$ changes (calculating $\frac{\Delta x}{\Delta y}$ instead of $\frac{\Delta y}{\Delta x}$)
Failing to rearrange the final equation into the form $ax + by + c = 0$ when explicitly instructed, resulting in a loss of the final accuracy mark
Confusing the coordinates of the circle's center with the coordinates of the point of contact when setting up the line equation

Study Guide Available

Comprehensive revision notes & examples

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Key Terminology

Essential terms to know

Perpendicularity of radius and tangent

Coordinate geometry of the circle

Point-gradient form of linear equations

Discriminant conditions for tangency

Likely Command Words

How questions on this topic are typically asked

Find

Show that

Calculate

Determine

Work out

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