| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Range of parameter for intersection |
| Difficulty | Standard +0.3 Part (a) involves routine completing the square to find centre and radius, then simple arithmetic for the lowest point. Part (b) requires substituting the line into the circle equation and using the discriminant condition, which is a standard technique but involves algebraic manipulation across multiple steps. Overall slightly above average difficulty due to the discriminant work in part (b). |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Express as \((x+3)^2 + (y-1)^2 = 26+9+1 = 36\) | M1 | Completing the square on \(x\) and \(y\) or using the form \(x^2+y^2+2gx+2fy+c=0\), centre \((-g,-f)\) and radius \(\sqrt{g^2+f^2-c}\). SOI by correct answer. |
| Centre \((-3, 1)\) | B1 | |
| Radius \(6\) | B1 | |
| Lowest point is \((-3, -5)\) | A1 FT | FT on *their* centre and *their* radius. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Intersects when \(x^2+(kx-5)^2+6x-2(kx-5)-26=0\) or \((x+3)^2+(kx-5-1)^2=36\) | *M1 | Substituting \(y=kx-5\) into *their* circle equation or rearranging and equating \(y\). |
| \(x^2+k^2x^2-10kx+25+6x-2kx+10-26=0\) or \(x^2+6x+9+k^2x^2-12kx+36=36\) leading to \((k^2+1)x^2+(6-12k)x+9=0\) | DM1, A1 | Rearranging to 3-term quadratic (terms grouped, all on one side). Allow 1 error. Correct quadratic (need to see 9 as constant term). |
| \((6-12k)^2 - 4(k^2+1)\times 9 > 0\), leading to \(144k^2-144k+36-36k^2-36>0\) | DM1 | Using discriminant \(b^2-4ac>0\) with *their* values. Allow if in square root. |
| \(108k^2-144k=0\) leading to \(k=0\) or \(k=\frac{4}{3}\) | A1 | Need not see method for solving. |
| \(k<0,\ k>\frac{4}{3}\) | A1 | Do not accept \(\frac{4}{3} |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Express as $(x+3)^2 + (y-1)^2 = 26+9+1 = 36$ | M1 | Completing the square on $x$ and $y$ or using the form $x^2+y^2+2gx+2fy+c=0$, centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$. SOI by correct answer. |
| Centre $(-3, 1)$ | B1 | |
| Radius $6$ | B1 | |
| Lowest point is $(-3, -5)$ | A1 FT | FT on *their* centre and *their* radius. |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Intersects when $x^2+(kx-5)^2+6x-2(kx-5)-26=0$ **or** $(x+3)^2+(kx-5-1)^2=36$ | *M1 | Substituting $y=kx-5$ into *their* circle equation or rearranging and equating $y$. |
| $x^2+k^2x^2-10kx+25+6x-2kx+10-26=0$ **or** $x^2+6x+9+k^2x^2-12kx+36=36$ leading to $(k^2+1)x^2+(6-12k)x+9=0$ | DM1, A1 | Rearranging to 3-term quadratic (terms grouped, all on one side). Allow 1 error. Correct quadratic (need to see 9 as constant term). |
| $(6-12k)^2 - 4(k^2+1)\times 9 > 0$, leading to $144k^2-144k+36-36k^2-36>0$ | DM1 | Using discriminant $b^2-4ac>0$ with *their* values. Allow if in square root. |
| $108k^2-144k=0$ leading to $k=0$ or $k=\frac{4}{3}$ | A1 | Need not see method for solving. |
| $k<0,\ k>\frac{4}{3}$ | A1 | Do not accept $\frac{4}{3}<k<0$. |
---9 The equation of a circle is $x ^ { 2 } + y ^ { 2 } + 6 x - 2 y - 26 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of the circle and the radius. Hence find the coordinates of the lowest point on the circle.
\item Find the set of values of the constant $k$ for which the line with equation $y = k x - 5$ intersects the circle at two distinct points.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q9 [10]}}