| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation involving finding the point of tangency |
| Difficulty | Standard +0.3 This is a standard multi-part circle question requiring: (a) finding circle equation from center and point (routine), (b) finding tangent using perpendicular radius (standard technique), and (c) finding point of tangency given tangent equation (slightly more involved but still algorithmic). All parts use well-practiced methods with no novel insight required, 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/Working | Mark | Guidance |
| See \((x \pm 1)^2 + (y \pm 3)^2 = r^2\) or \(x^2 + y^2 \pm 2x \pm 6y + c = 0\) | M1 | Centre at \((1,-3)\) |
| Attempt \(\sqrt{(8-1)^2+(-2-(-3))^2}\) or \((8-1)^2+(-2-(-3))^2\) | M1 | Attempt at finding radius or radius² |
| \((x-1)^2 + (y+3)^2 = 50\) or \(x^2 + y^2 - 2x + 6y - 40 = 0\) | A1, A1 | Accept either form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of \(AP = \frac{1}{7}\) | B1 | Implied by use of \(-7\) in tangent |
| Gradient of tangent is \(-7\) | M1 | Uses negative reciprocal |
| Equation of tangent: \((y+2) = -7(x-8)\) | dM1 | Linear equation through \((8,-2)\) with negative reciprocal gradient |
| \(y = -7x + 54\) or \(m=-7\), \(c=54\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x+6\) meets circle: \((x-1)^2+(x+9)^2=50\) or \(2x^2+16x+32=0\) | M1 | Eliminates \(x\) or \(y\) from two relevant equations |
| Correct quadratic in \(x\) or \(y\): \(2x^2+16x+32=0\) or \(2y^2-8y+8=0\) | A1 | Correct quadratic |
| Solve to give \(x\) or \(y\) | M1 | Usual rules; first M must have been scored |
| Substitute to give second coordinate | dM1 | Dependent on both previous M marks |
| \((-4,\ 2)\) only | A1 | Accept \(x=-4\), \(y=2\); withhold if two answers given |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| See $(x \pm 1)^2 + (y \pm 3)^2 = r^2$ or $x^2 + y^2 \pm 2x \pm 6y + c = 0$ | M1 | Centre at $(1,-3)$ |
| Attempt $\sqrt{(8-1)^2+(-2-(-3))^2}$ or $(8-1)^2+(-2-(-3))^2$ | M1 | Attempt at finding radius or radius² |
| $(x-1)^2 + (y+3)^2 = 50$ or $x^2 + y^2 - 2x + 6y - 40 = 0$ | A1, A1 | Accept either form |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $AP = \frac{1}{7}$ | B1 | Implied by use of $-7$ in tangent |
| Gradient of tangent is $-7$ | M1 | Uses negative reciprocal |
| Equation of tangent: $(y+2) = -7(x-8)$ | dM1 | Linear equation through $(8,-2)$ with negative reciprocal gradient |
| $y = -7x + 54$ or $m=-7$, $c=54$ | A1 | cao |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x+6$ meets circle: $(x-1)^2+(x+9)^2=50$ or $2x^2+16x+32=0$ | M1 | Eliminates $x$ or $y$ from two relevant equations |
| Correct quadratic in $x$ or $y$: $2x^2+16x+32=0$ or $2y^2-8y+8=0$ | A1 | Correct quadratic |
| Solve to give $x$ or $y$ | M1 | Usual rules; first M must have been scored |
| Substitute to give second coordinate | dM1 | Dependent on both previous M marks |
| $(-4,\ 2)$ only | A1 | Accept $x=-4$, $y=2$; withhold if two answers given |
---13. The circle $C$ has centre $A ( 1 , - 3 )$ and passes through the point $P ( 8 , - 2 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the circle $C$.
The line $l _ { 1 }$ is the tangent to $C$ at the point $P$.
\item Find an equation for $l _ { 1 }$, giving your answer in the form $y = m x + c$
The line $l _ { 2 }$, with equation $y = x + 6$, is the tangent to $C$ at the point $Q$.
\item Find the coordinates of the point $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q13 [13]}}