| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard circle techniques: distance formula for radius, circle equation form, and perpendicular gradient for tangent. All steps are routine applications of well-practiced methods with no problem-solving insight required, making it easier than average but not trivial due to the coordinate manipulation involved. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{4-2x}{2y-10}\) e.g. \(2(x-2)+2(y-5)\frac{dy}{dx}=0\) or e.g. \(y=(100-(x-2)^2)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = -(x-2)(100-(x-2)^2)^{-\frac{1}{2}}\) | B1 | May be implied by \(\frac{3}{4}\) |
| Attempts to find gradient of tangent at \(P\) using \(\frac{dy}{dx}\) and given coordinates | M1 | Do not be concerned with poorly differentiated equations; just look for \(x\) (and possibly \(y\)) coordinates substituted into differential equation leading to gradient value. Condone sign slips. |
| As above in main scheme | M1 | |
| As above in main scheme | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x-2)(x_1-2)+(y-5)(y_1-5)\)... | B1 | May be implied by further work |
| \((x-2)(x_1-2)+(y-5)(y_1-5)=\text{"100"}\) | M1 | Sets equal to their radius |
| \((x-2)(8-2)+(y-5)(-3-5)...=\text{"100"}\) | M1 | Uses point \(P\) and sets equal to radius |
| \(3x-4y-36=0\) | A1 | All coefficients integers, all terms on one side |
# Question 3 (Circle Tangent - Alt methods):
## Alternative method using differentiation:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{4-2x}{2y-10}$ e.g. $2(x-2)+2(y-5)\frac{dy}{dx}=0$ or e.g. $y=(100-(x-2)^2)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = -(x-2)(100-(x-2)^2)^{-\frac{1}{2}}$ | B1 | May be implied by $\frac{3}{4}$ |
| Attempts to find gradient of tangent at $P$ using $\frac{dy}{dx}$ and given coordinates | M1 | Do not be concerned with poorly differentiated equations; just look for $x$ (and possibly $y$) coordinates substituted into differential equation leading to gradient value. Condone sign slips. |
| As above in main scheme | M1 | |
| As above in main scheme | A1 | |
## Alternative method using tangent formula:
Using $(x-a)(x_1-a)+(y-b)(y_1-b)=r^2$ with centre $(a,b)$ and point $(x_1, y_1)$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x-2)(x_1-2)+(y-5)(y_1-5)$... | B1 | May be implied by further work |
| $(x-2)(x_1-2)+(y-5)(y_1-5)=\text{"100"}$ | M1 | Sets equal to their radius |
| $(x-2)(8-2)+(y-5)(-3-5)...=\text{"100"}$ | M1 | Uses point $P$ and sets equal to radius |
| $3x-4y-36=0$ | A1 | All coefficients integers, all terms on one side |
---\begin{enumerate}
\item A circle $C$ has centre $( 2,5 )$
\end{enumerate}
Given that the point $P ( 8 , - 3 )$ lies on $C$\\
(a) (i) find the radius of $C$\\
(ii) find an equation for $C$\\
(b) Find the equation of the tangent to $C$ at $P$ giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers to be found.
\hfill \mbox{\textit{Edexcel P2 2023 Q3 [7]}}