| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring standard techniques: substituting points into the circle equation to find unknowns (forming simultaneous equations), then finding a tangent using the radius-perpendicular property. Both parts are routine applications of circle theory with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.03d 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1+1+a+b-12=0 \Rightarrow a+b=10\) \(4+36+2a-6b-12=0 \Rightarrow 2a-6b=-28\) | B1 B1 | B1 for each equation. Allow unsimplified. Can be implied by correct values for \(a\) and \(b\) |
| \(a=4, b=6\) | B1 | |
| Centre is \(\left(-\frac{\text{their }a}{2}, -\frac{\text{their }b}{2}\right)\) \([-2,-3]\) | B1 FT | Or \(x=-2, y=-3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of \(AC\) is \(\frac{1-\text{their }y}{1-\text{their }x}\left[=\frac{1--3}{1--2}=\frac{1+3}{1+2}=\frac{4}{3}\right]\) | *M1 | Using their centre correctly |
| Gradient of tangent \(= \frac{-1}{\text{their }\frac{4}{3}}\left[=-\frac{3}{4}\right]\) | A1 FT | Use of \(m_1m_2=-1\) to obtain gradient of tangent |
| Equation: \(y-1=\text{'their}-\frac{3}{4}\text{'}(x-1)\) or \(y=-\frac{3}{4}x+\frac{7}{4}\) | DM1 | Using \((1,1)\) with their gradient of tangent at \(A\) |
| \(3x+4y=7\) or \(4y+3x=7\) or integer multiples of these | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2x+2y\frac{dy}{dx}+4+6\frac{dy}{dx}=0\) | *M1 | Implicit differentiation with at least one \(y\) term differentiated correctly |
| \(8\frac{dy}{dx}=-6 \Rightarrow \frac{dy}{dx}=-\frac{6}{8}\) | A1 | |
| Equation: \(y-1=\text{'their}-\frac{3}{4}\text{'}(x-1)\) or \(y=-\frac{3}{4}x+\frac{7}{4}\) | DM1 | Using \((1,1)\) with their gradient of tangent at \(A\) |
| \(3x+4y=7\) or \(4y+3x=7\) or integer multiples of these | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx}=\frac{1}{2}\{25-(x+2)^2\}^{-\frac{1}{2}}(-2x-4)\) | *M1 | Rearranging to form \(y=\) and differentiating using the chain rule |
| \(\frac{dy}{dx}=\frac{1}{2}(25-9)^{-\frac{1}{2}}(-6)=-\frac{6}{8}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equation: \(y - 1 = \text{'their'} -\frac{3}{4}(x-1)\) or \(y = -\frac{3}{4}x + \frac{7}{4}\) | DM1 | Using \((1,1)\) with *their* gradient of the tangent at \(A\) |
| \(3x + 4y = 7\) or \(4y + 3x = 7\), or integer multiples of these | A1 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1+1+a+b-12=0 \Rightarrow a+b=10$ $4+36+2a-6b-12=0 \Rightarrow 2a-6b=-28$ | B1 B1 | B1 for each equation. Allow unsimplified. Can be implied by correct values for $a$ and $b$ |
| $a=4, b=6$ | B1 | |
| Centre is $\left(-\frac{\text{their }a}{2}, -\frac{\text{their }b}{2}\right)$ $[-2,-3]$ | B1 FT | Or $x=-2, y=-3$ |
---
## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of $AC$ is $\frac{1-\text{their }y}{1-\text{their }x}\left[=\frac{1--3}{1--2}=\frac{1+3}{1+2}=\frac{4}{3}\right]$ | *M1 | Using their centre correctly |
| Gradient of tangent $= \frac{-1}{\text{their }\frac{4}{3}}\left[=-\frac{3}{4}\right]$ | A1 FT | Use of $m_1m_2=-1$ to obtain gradient of tangent |
| Equation: $y-1=\text{'their}-\frac{3}{4}\text{'}(x-1)$ or $y=-\frac{3}{4}x+\frac{7}{4}$ | DM1 | Using $(1,1)$ with their gradient of tangent at $A$ |
| $3x+4y=7$ or $4y+3x=7$ or integer multiples of these | A1 | |
**Alternative method for 8(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $2x+2y\frac{dy}{dx}+4+6\frac{dy}{dx}=0$ | *M1 | Implicit differentiation with at least one $y$ term differentiated correctly |
| $8\frac{dy}{dx}=-6 \Rightarrow \frac{dy}{dx}=-\frac{6}{8}$ | A1 | |
| Equation: $y-1=\text{'their}-\frac{3}{4}\text{'}(x-1)$ or $y=-\frac{3}{4}x+\frac{7}{4}$ | DM1 | Using $(1,1)$ with their gradient of tangent at $A$ |
| $3x+4y=7$ or $4y+3x=7$ or integer multiples of these | A1 | |
**Alternative method for 8(b) (rearranging):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx}=\frac{1}{2}\{25-(x+2)^2\}^{-\frac{1}{2}}(-2x-4)$ | *M1 | Rearranging to form $y=$ and differentiating using the chain rule |
| $\frac{dy}{dx}=\frac{1}{2}(25-9)^{-\frac{1}{2}}(-6)=-\frac{6}{8}$ | A1 | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equation: $y - 1 = \text{'their'} -\frac{3}{4}(x-1)$ or $y = -\frac{3}{4}x + \frac{7}{4}$ | DM1 | Using $(1,1)$ with *their* gradient of the tangent at $A$ |
| $3x + 4y = 7$ or $4y + 3x = 7$, or integer multiples of these | A1 | |
---8 The equation of a circle is $x ^ { 2 } + y ^ { 2 } + a x + b y - 12 = 0$. The points $A ( 1,1 )$ and $B ( 2 , - 6 )$ lie on the circle.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$ and hence find the coordinates of the centre of the circle.
\item Find the equation of the tangent to the circle at the point $A$, giving your answer in the form $p x + q y = k$, where $p , q$ and $k$ are integers.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q8 [8]}}