| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.3 This is a straightforward circle question requiring standard techniques: finding the center (midpoint of diameter), radius (distance formula), circle equation, and tangent (perpendicular to radius). All steps are routine applications of formulas with no conceptual challenges, making it slightly easier than average for A-level. |
| Spec | 1.03b 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/Working | Mark | Guidance |
| Mid-point of \(AB = (2, -3)\) | M1 A1 | M1: uses midpoint formula |
| \(r^2 = (12-"2")^2 + (2-"-3")^2\) | M1 | Finds radius using distance from centre to a point |
| \(r^2 = 125\) | A1 | Accept \(r = \sqrt{125}\) |
| \(125 = (x-2)^2 + (y+3)^2\) | M1 A1 | M1: attempt correct circle equation form; A1: correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient from \((2,-3)\) to \((4,8)\) \(= \frac{8-"-3"}{4-"-2"} = \frac{11}{2}\) | M1 | Uses centre and \((4,8)\); gradient incorrect with no method shown \(\Rightarrow\) M0 |
| \(ZM\) has gradient \(-\frac{1}{m} = -\frac{2}{11}\) | M1 | Finds negative reciprocal |
| \(y - 8 = -\frac{2}{11}(x-4)\) | ddM1 | Correct line method with \((4,8)\) and perpendicular gradient; dependent on both M marks |
| \(2x + 11y - 96 = 0\) | A1 | cao — accept multiples; integer coefficients not required |
# Question 15:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mid-point of $AB = (2, -3)$ | M1 A1 | M1: uses midpoint formula |
| $r^2 = (12-"2")^2 + (2-"-3")^2$ | M1 | Finds radius using distance from centre to a point |
| $r^2 = 125$ | A1 | Accept $r = \sqrt{125}$ |
| $125 = (x-2)^2 + (y+3)^2$ | M1 A1 | M1: attempt correct circle equation form; A1: correct answer |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient from $(2,-3)$ to $(4,8)$ $= \frac{8-"-3"}{4-"-2"} = \frac{11}{2}$ | M1 | Uses centre and $(4,8)$; gradient incorrect with no method shown $\Rightarrow$ M0 |
| $ZM$ has gradient $-\frac{1}{m} = -\frac{2}{11}$ | M1 | Finds negative reciprocal |
| $y - 8 = -\frac{2}{11}(x-4)$ | ddM1 | Correct line method with $(4,8)$ and perpendicular gradient; dependent on both M marks |
| $2x + 11y - 96 = 0$ | A1 | cao — accept multiples; integer coefficients not required |
---15. The points $A$ and $B$ have coordinates $( - 8 , - 8 )$ and $( 12,2 )$ respectively. $A B$ is the diameter of a circle $C$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the circle $C$.
The point $( 4,8 )$ also lies on $C$.
\item Find an equation of the tangent to $C$ at the point ( 4,8 ), giving your answer in the form $a x + b y + c = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q15 [10]}}