| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using right angle in semicircle |
| Difficulty | Moderate -0.3 This is a structured multi-part question with clear signposting ('show', 'hence') that guides students through standard techniques: verifying perpendicularity via gradients, using the right-angle-in-semicircle property to locate the center, finding circle equation, and determining a tangent. While it requires multiple steps, each part is routine application of well-practiced methods with no novel problem-solving required. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient of \(AB = -\frac{3}{5}\), gradient of \(BC = \frac{5}{3}\) or lengths of all 3 sides or vectors | M1 | Attempting to find required gradients, sides or vectors |
| \(m_{ab}m_{bc} = -1\) or Pythagoras or \(\overrightarrow{AB}.\overrightarrow{BC} = 0\) or \(\cos ABC = 0\) from cosine rule | A1 | WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Centre = mid-point of \(AC = (2,4)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x - \text{their } x_c)^2 + (y - \text{their } y_c)^2 [= r^2]\) or \((\text{their } x_c - x)^2 + (\text{their } y_c - y)^2 = [r^2]\) | M1 | Use of circle equation with *their* centre |
| \((x-2)^2 + (y-4)^2 = 17\) | A1 | Accept \(x^2 - 4x + y^2 - 8y + 3 = 0\) OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{x+3}{2}, \frac{y+0}{2}\right) = (2,4)\) or \(\mathbf{BE} = 2\mathbf{BD} = 2\begin{pmatrix}-1\\4\end{pmatrix}\); or equation of \(BE\) is \(y = -4(x-3)\) or \(y-4=-4(x-2)\) leading to \(y = -4x+12\); substitute equation of \(BE\) into circle and form 3-term quadratic | M1 | Use of mid-point formula, vectors, steps on a diagram. May be seen to find \(x\) coordinate at \(E\) |
| \((x,y) = (1,8)\) or \(\mathbf{OE} = \begin{pmatrix}3\\0\end{pmatrix} + \begin{pmatrix}-2\\8\end{pmatrix} = \begin{pmatrix}1\\8\end{pmatrix}\) | A1 | \(E = (1,8)\). Accept without working for both marks SC B2 |
| Gradient of \(BD\), \(m = -4\) or gradient \(AC = \frac{1}{4}\) = gradient of tangent | B1 | Or gradient of \(BE = -4\) |
| Equation of tangent is \(y - 8 = \frac{1}{4}(x-1)\) OE | M1 A1 | For M1, equation through *their* \(E\) or \((1,8)\) (not \(A\), \(B\) or \(C\)) and with gradient \(\frac{-1}{\text{their}-4}\) |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of $AB = -\frac{3}{5}$, gradient of $BC = \frac{5}{3}$ or lengths of all 3 sides or vectors | M1 | Attempting to find required gradients, sides or vectors |
| $m_{ab}m_{bc} = -1$ or Pythagoras or $\overrightarrow{AB}.\overrightarrow{BC} = 0$ or $\cos ABC = 0$ from cosine rule | A1 | WWW |
---
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre = mid-point of $AC = (2,4)$ | B1 | |
---
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x - \text{their } x_c)^2 + (y - \text{their } y_c)^2 [= r^2]$ or $(\text{their } x_c - x)^2 + (\text{their } y_c - y)^2 = [r^2]$ | M1 | Use of circle equation with *their* centre |
| $(x-2)^2 + (y-4)^2 = 17$ | A1 | Accept $x^2 - 4x + y^2 - 8y + 3 = 0$ OE |
---
## Question 10(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{x+3}{2}, \frac{y+0}{2}\right) = (2,4)$ or $\mathbf{BE} = 2\mathbf{BD} = 2\begin{pmatrix}-1\\4\end{pmatrix}$; or equation of $BE$ is $y = -4(x-3)$ or $y-4=-4(x-2)$ leading to $y = -4x+12$; substitute equation of $BE$ into circle and form 3-term quadratic | M1 | Use of mid-point formula, vectors, steps on a diagram. May be seen to find $x$ coordinate at $E$ |
| $(x,y) = (1,8)$ or $\mathbf{OE} = \begin{pmatrix}3\\0\end{pmatrix} + \begin{pmatrix}-2\\8\end{pmatrix} = \begin{pmatrix}1\\8\end{pmatrix}$ | A1 | $E = (1,8)$. Accept without working for both marks **SC B2** |
| Gradient of $BD$, $m = -4$ or gradient $AC = \frac{1}{4}$ = gradient of tangent | B1 | Or gradient of $BE = -4$ |
| Equation of tangent is $y - 8 = \frac{1}{4}(x-1)$ OE | M1 A1 | For M1, equation through *their* $E$ or $(1,8)$ (not $A$, $B$ or $C$) and with gradient $\frac{-1}{\text{their}-4}$ |
---10 Points $A ( - 2,3 ) , B ( 3,0 )$ and $C ( 6,5 )$ lie on the circumference of a circle with centre $D$.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $A B C = 90 ^ { \circ }$.
\item Hence state the coordinates of $D$.
\item Find an equation of the circle.\\
The point $E$ lies on the circumference of the circle such that $B E$ is a diameter.
\item Find an equation of the tangent to the circle at $E$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q10 [10]}}