WJEC Unit 1 Specimen — Question 1 7 marks

Exam BoardWJEC
ModuleUnit 1 (Unit 1)
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeTangent equation at a known point on circle
DifficultyModerate -0.8 This is a straightforward circle question requiring completion of the square to find centre and radius (standard technique), then using the perpendicular gradient property for the tangent. Both parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple steps and careful algebra.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations

The circle \(C\) has centre \(A\) and equation $$x^2 + y^2 - 2x + 6y - 15 = 0.$$
  1. Find the coordinates of \(A\) and the radius of \(C\). [3]
  2. The point \(P\) has coordinates \((4, -7)\) and lies on \(C\). Find the equation of the tangent to \(C\) at \(P\). [4]

AnswerMarks Guidance
(a) A correct method for finding the radius, e.g., trying to rewrite the equation of the circle in the form \((x – a)^2 + (y – b)^2 = r^2\)B1, M1, A1 AO1
Radius = 5A1 AO1
(b) Gradient \(AP = \frac{\text{increase in } y}{\text{increase in } x}\)M1 AO1
Gradient \(AP = \frac{(-7) – (-3)}{4 – 1} = -\frac{4}{3}\)A1 AO1 (f.t. candidate's coordinates for A)
Use of \(m_{\text{tan}} \times m_{\text{rad}} = -1\)M1 AO1
Equation of tangent is: \(y – (-7) = \frac{3}{4}(x – 4)\)A1 AO1 (f.t. candidate's gradient for AP)
Total: [7]
**(a)** A correct method for finding the radius, e.g., trying to rewrite the equation of the circle in the form $(x – a)^2 + (y – b)^2 = r^2$ | B1, M1, A1 | AO1
Radius = 5 | A1 | AO1

**(b)** Gradient $AP = \frac{\text{increase in } y}{\text{increase in } x}$ | M1 | AO1
Gradient $AP = \frac{(-7) – (-3)}{4 – 1} = -\frac{4}{3}$ | A1 | AO1 (f.t. candidate's coordinates for A)
Use of $m_{\text{tan}} \times m_{\text{rad}} = -1$ | M1 | AO1
Equation of tangent is: $y – (-7) = \frac{3}{4}(x – 4)$ | A1 | AO1 (f.t. candidate's gradient for AP)

**Total: [7]**

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The circle $C$ has centre $A$ and equation
$$x^2 + y^2 - 2x + 6y - 15 = 0.$$

\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$ and the radius of $C$. [3]

\item The point $P$ has coordinates $(4, -7)$ and lies on $C$. Find the equation of the tangent to $C$ at $P$. [4]
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 1  Q1 [7]}}