Moderate -0.8 This is a straightforward multi-part circle question requiring completing the square to find centre and radius, substituting x=0 for y-intercepts, and finding a tangent using the perpendicular radius method. All parts are standard C2 techniques with no problem-solving insight required, making it easier than average but not trivial due to the multiple steps involved.
the \(y\) coordinates of the points where the circle \(C\) crosses the \(y\)-axis.
Find an equation of the tangent to \(C\) at the point ( 2,0 ), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
You may mark (a) and (b) together: \(x^2 + y^2 - 2x + 14y = 0\). Obtain LHS as \((x-1)^2 + (y+7)^2 = ...\)
M1 A1
Centre is (1, –7)
(2)
(b)
Uses \(r^2 = a^2 + b^2\) or \(r = \sqrt{a^2 + b^2}\) where their centre was at (\(\pm a, \pm b\)). \(r = \sqrt{50}\) or \(5\sqrt{2}\)
M1 A1
(2)
(c)
Substitute \(x = 0\) in either form of equation of circle and solve resulting quadratic to give \(y =\)
M1 A1
\(y^2 + 14y = 0\) so \(y = 0\) and \(-14\) or \((y \pm 7)^2 - 49 = 0\) so \(y = 0\) and \(-14\)
(2)
(d)
Gradient of radius joining centre to (2,0) is \(\frac{"-7"- 0}{"
1"- 2} = (-7)\)
Gradient of tangent is \(\frac{-1}{m} (= -\frac{1}{7})\)
So equation is \(y - 0 = -\frac{1}{7}(x - 2)\) and so \(x + 7y - 2 = 0\)
M1, A1
M1: Line equation through (2, 0) with changed gradient so if use \(y = mx + c\) they need to use (2, 0) to find \(c\). A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
(4)
(10 marks)
Alternative Methods which may be seen:
Answer
Marks
Guidance
Part
Answer/Working
Marks
(a)
Method 2: Comparing with \(x^2 + y^2 + 2gx + 2fy + c = 0\) to write down centre \((–g, –f)\) directly. Condone sign errors for this M mark. Centre is (1, –7).
M1 A1
(b)
Method 2: Using \(\sqrt{g^2 + f^2 - c}\). So \(r = \sqrt{50}\) or \(5\sqrt{2}\)
M1 A1
(d)
Method 3: Using Implicit Differentiation: \(2x + 2y\frac{dy}{dx} - 2 + 14\frac{dy}{dx} = 0\) or \(2(x-1) + 2(y+7)\frac{dy}{dx} = 0\)
M1
\(\frac{dy}{dx}... (\frac{2-2x}{14+2y}\) or \(\frac{-2}{14})\)
M1
M1: Rearranges their differentiated expression and substitutes \(x=2, y=0\) to obtain gradient – allow slips. (It should be \(\frac{dy}{dx} = \frac{14+2y}{14}\)) If there is no \(y\) term this mark may be earned for substitution of \(x=2\) as \(y = 0\) is not needed.
So equation is \(y - 0 = -\frac{1}{7}(x - 2)\) and so \(x + 7y - 2 = 0\)
M1, A1
M1: Line equation through (2, 0) with their obtained gradient so if use \(y = mx + c\) need use (2, 0) to find \(c\). A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
(4)
(d)
Method 4: Making \(y\) the subject of formula and differentiating: \(y = -7 \pm \sqrt{150 - (x-1)^2}\) so \(\frac{dy}{dx} = \pm \frac{1}{2} \times -2(x-1)\{50-(x-1)^2\}^{-\frac{1}{2}}\)
M1
At \(x=2\), \(\frac{dy}{dx} = \pm \frac{1}{7}\)
M1 (contd next page)
Chooses \(\frac{dy}{dx} = -\frac{1}{7}\) and so \(x + 7y - 2 = 0\)
A1
Notes:
(a) M1: as in scheme and can be implied by (\(\pm1, \pm7\)) even if this follows some poor working. A1: (1, –7)
(b) M1: Uses \(r^2 = a^2 + b^2\) or \(r = \sqrt{a^2 + b^2}\) where their centre was at (\(\pm a, \pm b\)). A1: \(\sqrt{50}\) or \(5\sqrt{2}\). not 50 only.
Answer
Marks
Guidance
Special case: if centre given as (–1, –7) or (1, 7) or (–1, 7) or coordinates given wrong way round – allow M1A1 for \(r = 5\sqrt{2}\) worked correctly. \(r^2 = "
1
"+"
(c) M1: As in scheme – allow for just one value of \(y\). A1: Accept (0, 0), (0, –14) or \(y = 0, y = -14\) or just 0 and –14
(d) Method 1:
M1: Correct method for gradient – if no method shown answer must be correct to earn this mark. If \(x\) and \(y\) coordinates are confused and fraction is upside down this is M0 even if formula is quoted as there is no evidence of understanding.
M1: Correct negative reciprocal of their gradient
M1: Line equation through (2, 0) with changed gradient so if use \(y = mx + c\) need use (2, 0) to find \(c\)
A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
(d) Method 3:
M1: Correct implicit differentiation (no errors)
M1: Rearranges their differentiated expression and substitutes \(x=2, y=0\) to obtain gradient – allow slips. (It should be \(\frac{dy}{dx} = \frac{14+2y}{14}\)). If there is no \(y\) term this mark may be earned for substitution of \(x=2\) as \(y = 0\) is not needed
M1: Line equation through (2, 0) with their obtained gradient so if use \(y = mx + c\) need use (2, 0) to find \(c\)
A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
(d) Method 4:
M1: Correct rearrangement and differentiation (no errors)
M1: Substitutes \(x=2\) to obtain gradient – allow minus and plus
M1: Line equation through (2, 0) with their obtained gradient so if use \(y = mx + c\) need use (2, 0) to find \(c\)
A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
| **Part** | **Answer/Working** | **Marks** | **Notes** |
|---|---|---|---|
| (a) | You may mark (a) and (b) together: $x^2 + y^2 - 2x + 14y = 0$. Obtain LHS as $(x-1)^2 + (y+7)^2 = ...$ | M1 A1 | M1: as in scheme and can be implied by ($\pm1, \pm7$) even if this follows some poor working. A1: (1, –7) |
| | Centre is (1, –7) | | |
| | | **(2)** | |
| (b) | Uses $r^2 = a^2 + b^2$ or $r = \sqrt{a^2 + b^2}$ where their centre was at ($\pm a, \pm b$). $r = \sqrt{50}$ or $5\sqrt{2}$ | M1 A1 | M1: Uses $r^2 = a^2 + b^2$ or $r = \sqrt{a^2 + b^2}$ where their centre was at ($\pm a, \pm b$). A1: $\sqrt{50}$ or $5\sqrt{2}$. not 50 only. Special case: if centre given as (–1, –7) or (1, 7) or (–1, 7) or coordinates given wrong way round – allow M1A1 for $r = 5\sqrt{2}$ worked correctly. $r^2 = "|1|"+"|49|"$ If they get $r = 5\sqrt{2}$ after wrong statements such as $r^2 = "-1"+"49"$ then this is M0A0. $r = 5\sqrt{2}$ with no working earns M1A1 as there is no wrong work. |
| | | **(2)** | |
| (c) | Substitute $x = 0$ in either form of equation of circle and solve resulting quadratic to give $y =$ | M1 A1 | M1: As in scheme – allow for just one value of $y$. A1: Accept (0, 0), (0, –14) or $y = 0, y = -14$ or just 0 and –14 |
| | $y^2 + 14y = 0$ so $y = 0$ and $-14$ or $(y \pm 7)^2 - 49 = 0$ so $y = 0$ and $-14$ | | |
| | | **(2)** | |
| (d) | Gradient of radius joining centre to (2,0) is $\frac{"-7"- 0}{"|1"- 2} = (-7)$ | M1 M1 | M1: Correct method for gradient – if no method shown answer must be correct to earn this mark. If $x$ and $y$ coordinates are confused and fraction is upside down this is M0 even if formula is quoted as there is no evidence of understanding. M1: Correct negative reciprocal of their gradient. |
| | Gradient of tangent is $\frac{-1}{m} (= -\frac{1}{7})$ | | |
| | So equation is $y - 0 = -\frac{1}{7}(x - 2)$ and so $x + 7y - 2 = 0$ | M1, A1 | M1: Line equation through (2, 0) with changed gradient so if use $y = mx + c$ they need to use (2, 0) to find $c$. A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0) |
| | | **(4)** | |
| | | **(10 marks)** | |
**Alternative Methods which may be seen:**
| **Part** | **Answer/Working** | **Marks** | **Notes** |
|---|---|---|---|
| (a) | Method 2: Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$ to write down centre $(–g, –f)$ directly. Condone sign errors for this M mark. Centre is (1, –7). | M1 A1 | |
| (b) | Method 2: Using $\sqrt{g^2 + f^2 - c}$. So $r = \sqrt{50}$ or $5\sqrt{2}$ | M1 A1 | |
| (d) | Method 3: Using Implicit Differentiation: $2x + 2y\frac{dy}{dx} - 2 + 14\frac{dy}{dx} = 0$ or $2(x-1) + 2(y+7)\frac{dy}{dx} = 0$ | M1 | |
| | $\frac{dy}{dx}... (\frac{2-2x}{14+2y}$ or $\frac{-2}{14})$ | M1 | M1: Rearranges their differentiated expression and substitutes $x=2, y=0$ to obtain gradient – allow slips. (It should be $\frac{dy}{dx} = \frac{14+2y}{14}$) If there is no $y$ term this mark may be earned for substitution of $x=2$ as $y = 0$ is not needed. |
| | So equation is $y - 0 = -\frac{1}{7}(x - 2)$ and so $x + 7y - 2 = 0$ | M1, A1 | M1: Line equation through (2, 0) with their obtained gradient so if use $y = mx + c$ need use (2, 0) to find $c$. A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0) |
| | | **(4)** | |
| | | | |
| | | | |
| (d) | Method 4: Making $y$ the subject of formula and differentiating: $y = -7 \pm \sqrt{150 - (x-1)^2}$ so $\frac{dy}{dx} = \pm \frac{1}{2} \times -2(x-1)\{50-(x-1)^2\}^{-\frac{1}{2}}$ | M1 | |
| | At $x=2$, $\frac{dy}{dx} = \pm \frac{1}{7}$ | M1 (contd next page) | |
| | Chooses $\frac{dy}{dx} = -\frac{1}{7}$ and so $x + 7y - 2 = 0$ | A1 | |
**Notes:**
(a) M1: as in scheme and can be implied by ($\pm1, \pm7$) even if this follows some poor working. A1: (1, –7)
(b) M1: Uses $r^2 = a^2 + b^2$ or $r = \sqrt{a^2 + b^2}$ where their centre was at ($\pm a, \pm b$). A1: $\sqrt{50}$ or $5\sqrt{2}$. not 50 only.
Special case: if centre given as (–1, –7) or (1, 7) or (–1, 7) or coordinates given wrong way round – allow M1A1 for $r = 5\sqrt{2}$ worked correctly. $r^2 = "|1|"+"|49|"$. If get $r = 5\sqrt{2}$ after wrong statements such as $r^2 = "-1"+"49"$ then this is M0A0. $r = 5\sqrt{2}$ with no working earns M1A1 as there is no wrong work.
(c) M1: As in scheme – allow for just one value of $y$. A1: Accept (0, 0), (0, –14) or $y = 0, y = -14$ or just 0 and –14
(d) **Method 1:**
M1: Correct method for gradient – if no method shown answer must be correct to earn this mark. If $x$ and $y$ coordinates are confused and fraction is upside down this is M0 even if formula is quoted as there is no evidence of understanding.
M1: Correct negative reciprocal of their gradient
M1: Line equation through (2, 0) with changed gradient so if use $y = mx + c$ need use (2, 0) to find $c$
A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
(d) **Method 3:**
M1: Correct implicit differentiation (no errors)
M1: Rearranges their differentiated expression and substitutes $x=2, y=0$ to obtain gradient – allow slips. (It should be $\frac{dy}{dx} = \frac{14+2y}{14}$). If there is no $y$ term this mark may be earned for substitution of $x=2$ as $y = 0$ is not needed
M1: Line equation through (2, 0) with their obtained gradient so if use $y = mx + c$ need use (2, 0) to find $c$
A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
(d) **Method 4:**
M1: Correct rearrangement and differentiation (no errors)
M1: Substitutes $x=2$ to obtain gradient – allow minus and plus
M1: Line equation through (2, 0) with their obtained gradient so if use $y = mx + c$ need use (2, 0) to find $c$
A1: For any multiple of answer in scheme (The answer must be an equation so if "–0" is missing this is A0)
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\begin{enumerate}
\item The circle $C$ has equation
\end{enumerate}
$$x ^ { 2 } + y ^ { 2 } - 2 x + 14 y = 0$$
Find\\
(a) the coordinates of the centre of $C$,\\
(b) the exact value of the radius of $C$,\\
(c) the $y$ coordinates of the points where the circle $C$ crosses the $y$-axis.\\
(d) Find an equation of the tangent to $C$ at the point ( 2,0 ), giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
\hfill \mbox{\textit{Edexcel C2 2018 Q5 [10]}}