| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Standard +0.3 This is a straightforward two-part question on circle geometry. Part (a) requires finding a perpendicular bisector using midpoint and gradient (routine technique). Part (b) involves finding the circle center by solving simultaneous linear equations, then using the distance formula for the radius. All steps are standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mid-point is \((-1, 7)\) | B1 | |
| Gradient, \(m\), of \(AB\) is \(\frac{8}{12}\) | B1 | OE |
| \(y - 7 = -\frac{12}{8}(x+1)\) | M1 | |
| \(3x + 2y = 11\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Solve simultaneously \(12x - 5y = 70\) and \(3x + 2y = 11\) | M1 | using their \(3x+2y=11\) |
| \(x = 5,\ y = -2\) | A1 | |
| Attempt to find distance between their \((5,-2)\) and either \((-7,3)\) or \((5,11)\) | M1 | |
| \((r) = \sqrt{12^2 + 5^2}\) or \(\sqrt{13^2 + 0} = 13\) | A1 | |
| Equation of circle is \((x-5)^2 + (y+2)^2 = 169\) | A1 |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mid-point is $(-1, 7)$ | B1 | |
| Gradient, $m$, of $AB$ is $\frac{8}{12}$ | B1 | OE |
| $y - 7 = -\frac{12}{8}(x+1)$ | M1 | |
| $3x + 2y = 11$ | A1 | AG |
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## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Solve simultaneously $12x - 5y = 70$ and $3x + 2y = 11$ | M1 | using their $3x+2y=11$ |
| $x = 5,\ y = -2$ | A1 | |
| Attempt to find distance between their $(5,-2)$ and either $(-7,3)$ or $(5,11)$ | M1 | |
| $(r) = \sqrt{12^2 + 5^2}$ or $\sqrt{13^2 + 0} = 13$ | A1 | |
| Equation of circle is $(x-5)^2 + (y+2)^2 = 169$ | A1 | |
---10
\begin{enumerate}[label=(\alph*)]
\item The coordinates of two points $A$ and $B$ are $( - 7,3 )$ and $( 5,11 )$ respectively.\\
Show that the equation of the perpendicular bisector of $A B$ is $3 x + 2 y = 11$.
\item A circle passes through $A$ and $B$ and its centre lies on the line $12 x - 5 y = 70$.
Find an equation of the circle.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q10 [9]}}