| Exam Board | WJEC |
|---|---|
| Module | Unit 1 (Unit 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent from external point - intersection or geometric properties |
| Difficulty | Moderate -0.3 Part (a) is trivial recall (kite). Part (b) requires recognizing right angles at tangent points, using Pythagoras to find tangent length (√96 = 4√6), then calculating area as 2×(1/2×5×4√6) = 20√6. This is a standard coordinate geometry problem with straightforward steps, making it slightly easier than average but not trivial due to the multi-step calculation and surds. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (b) A kite | B1 | AO2 |
| A correct method for finding \(TR(TS)\) | M1 | AO3 |
| \(TR(TS) = \sqrt{96}\) | A1 | AO3 |
| Area \(OTR(OTS) = \frac{1}{2} \times \sqrt{96} \times 5\) | M1 | AO3 (f.t. candidate's derived value for \(TR(TS)\)) |
| m1 | AO3 | |
| Area \(OTRS = 2 \times\) Area \(OTR(OTS)\) | — | — |
| Area \(OTRS = 20\sqrt{6}\) | A1 | AO3 (c.a.o.) |
**(a)** (b) A kite | B1 | AO2
A correct method for finding $TR(TS)$ | M1 | AO3
$TR(TS) = \sqrt{96}$ | A1 | AO3
Area $OTR(OTS) = \frac{1}{2} \times \sqrt{96} \times 5$ | M1 | AO3 (f.t. candidate's derived value for $TR(TS)$)
m1 | AO3
Area $OTRS = 2 \times$ Area $OTR(OTS)$ | — | —
Area $OTRS = 20\sqrt{6}$ | A1 | AO3 (c.a.o.)
**Total: [6]**
---The circle $C$ has radius 5 and its centre is the origin.
The point $T$ has coordinates $(11, 0)$.
The tangents from $T$ to the circle $C$ touch $C$ at the points $R$ and $S$.
\begin{enumerate}[label=(\alph*)]
\item Write down the geometrical name for the quadrilateral $ORTS$. [1]
\item Find the exact value of the area of the quadrilateral $ORTS$. Give your answer in its simplest form. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 1 Q8 [6]}}