| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| 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 C2 circle question requiring standard techniques: finding radius using distance formula, writing circle equation in standard form, then finding tangent gradient (perpendicular to radius) and using point-slope form. All steps are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to multiple computational steps. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Way 1: \((x\mp2)^2+(y\pm1)^2=k,\ k>0\) | M1 | Uses centre to write equation of circle in correct form |
| Attempts \(r^2=(4-2)^2+(-5+1)^2\) | M1 | Attempts distance between two points to establish \(r^2\); allow one sign slip |
| \((x-2)^2+(y+1)^2=20\) | A1 | [3] |
| Way 2: \(x^2+y^2\mp4x\pm2y+c=0\) | M1 | |
| \(4^2+(-5)^2-4\times4+2\times-5+c=0\) | M1 | |
| \(x^2+y^2-4x+2y-15=0\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Gradient of radius from centre to \((4,-5)=-2\) | B1 | Must be correct; may be implied by following work |
| Tangent gradient \(=-\dfrac{1}{\text{their numerical gradient of radius}}\) | M1 | Uses negative reciprocal |
| Equation of tangent is \((y+5)=\frac{1}{2}(x-4)\) | M1 | Uses \(y-y_1=m(x-x_1)\) with \((4,-5)\) and changed gradient |
| \(x-2y-14=0\) (or \(2y-x+14=0\) or integer multiples) | A1 | Answers in scheme or multiples; must have "\(=0\)" [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Quotes \(xx'+yy'-2(x+x')+(y+y')-15=0\) and substitutes \((4,-5)\) | B1 | |
| \(4x-5y-2(x+4)+(y-5)-15=0\) so \(2x-4y-28=0\) | M1, M1, A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Either \(2(x-2)+2(y+1)\frac{dy}{dx}=0\) or states \(y=-1-\sqrt{20-(x-2)^2}\) so \(\frac{dy}{dx}=\frac{(x-2)}{\sqrt{20-(x-2)^2}}\) | B1 | Differentiation must be accurate |
| Substitute \(x=4,\ y=-5\) after valid differentiation | M1 | Must substitute into their gradient function |
| \((y+5)=\frac{1}{2}(x-4)\) so \(x-2y-14=0\) | M1 A1 | [4] |
# Question 2:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| **Way 1:** $(x\mp2)^2+(y\pm1)^2=k,\ k>0$ | M1 | Uses centre to write equation of circle in correct form |
| Attempts $r^2=(4-2)^2+(-5+1)^2$ | M1 | Attempts distance between two points to establish $r^2$; allow one sign slip |
| $(x-2)^2+(y+1)^2=20$ | A1 | **[3]** |
| **Way 2:** $x^2+y^2\mp4x\pm2y+c=0$ | M1 | |
| $4^2+(-5)^2-4\times4+2\times-5+c=0$ | M1 | |
| $x^2+y^2-4x+2y-15=0$ | A1 | **[3]** |
## Part (b) — Way 1
| Working/Answer | Mark | Guidance |
|---|---|---|
| Gradient of radius from centre to $(4,-5)=-2$ | B1 | Must be correct; may be implied by following work |
| Tangent gradient $=-\dfrac{1}{\text{their numerical gradient of radius}}$ | M1 | Uses negative reciprocal |
| Equation of tangent is $(y+5)=\frac{1}{2}(x-4)$ | M1 | Uses $y-y_1=m(x-x_1)$ with $(4,-5)$ and changed gradient |
| $x-2y-14=0$ (or $2y-x+14=0$ or integer multiples) | A1 | Answers in scheme or multiples; must have "$=0$" **[4]** |
**Way 2:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Quotes $xx'+yy'-2(x+x')+(y+y')-15=0$ and substitutes $(4,-5)$ | B1 | |
| $4x-5y-2(x+4)+(y-5)-15=0$ so $2x-4y-28=0$ | M1, M1, A1 | **[4]** |
**Way 3:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Either $2(x-2)+2(y+1)\frac{dy}{dx}=0$ **or** states $y=-1-\sqrt{20-(x-2)^2}$ so $\frac{dy}{dx}=\frac{(x-2)}{\sqrt{20-(x-2)^2}}$ | B1 | Differentiation must be accurate |
| Substitute $x=4,\ y=-5$ after valid differentiation | M1 | Must substitute into their gradient function |
| $(y+5)=\frac{1}{2}(x-4)$ so $x-2y-14=0$ | M1 A1 | **[4]** |
**Total: [7]**
---2. A circle $C$ with centre at the point $( 2 , - 1 )$ passes through the point $A$ at $( 4 , - 5 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the circle $C$.
\item Find an equation of the tangent to the circle $C$ at the point $A$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2015 Q2 [7]}}