Question 5 - A-Level Maths

OCR MEI C1 — Question 5 12 marks

Exam Board	OCR MEI
Module	C1 (Core Mathematics 1)
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Find parameter values for tangency using discriminant
Difficulty	Moderate -0.3 This is a standard multi-part circle question covering basic concepts: identifying center/radius from equation, finding intersections, substituting to form a quadratic, and using the discriminant condition for tangency. While it requires multiple steps and techniques, each part follows routine procedures taught in C1 with no novel problem-solving required, making it slightly easier than average.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 1.03e Complete the square: find centre and radius of circle

5 A circle has equation $( x - 2 ) ^ { 2 } + y ^ { 2 } = 20$.

Write down the radius of the circle and the coordinates of its centre.
Find the points of intersection of the circle with the $y$-axis and sketch the circle.
Show that, where the line $y = 2 x + k$ intersects the circle, $$5 x ^ { 2 } + ( 4 k - 4 ) x + k ^ { 2 } - 16 = 0$$
Hence find the values of $k$ for which the line $y = 2 x + k$ is a tangent to the circle.

Show mark scheme Show mark scheme source

Question 5(i):

Answer	Marks	Guidance
$\sqrt{20}$ isw or $2\sqrt{5}$	B1	0 for $\pm\sqrt{20}$
$(2, 0)$	B1

Total: [2]

Question 5(ii):

Answer	Marks	Guidance
subst of $x = 0$ into circle eqn soi	M1	or Pythagoras used on sketch of circle: $2^2 + y^2 = 20$ oe; M0 for just $y^2 = 20$; M1 for $y^2 = 16$ or for $y = 4$; ignore intersections with $x$-axis also found
$y = \pm 4$ oe	A1	or B2 for just $y = \pm 4$ seen oe; accept both 4 and $-4$ shown on $y$ axis on sketch if both values not stated
sketch of circle with centre $(2, 0)$ or ft their centre from (i)	B1	if the centre is not marked, it should look roughly correct by eye – coords need not be given on sketch; condone intersections with axes not marked; circle should intersect both $+$ve and $-$ve $x$- and $y$-axes; must be clear attempt at circle; ignore any tangents drawn

Total: [3]

Question 5(iii):

Answer	Marks	Guidance
$(x-2)^2 + (2x+k)^2 = 20$	M1	for attempt to subst $2x + k$ for $y$; allow for attempt to subst $k = y - 2x$ into given eqn
$x^2 - 4x + 4 + 4x^2 + 4kx + k^2 = 20$	M1 dep	for correct expansion of at least one set of brackets, dependent on first M1; similarly for those working backwards
$5x^2 + (4k-4)x + k^2 - 16 = 0$	A1	correct completion to given answer; dependent on both Ms; condone omission of further interim step if both sets of brackets expanded correctly, but for candidates working backwards, at least one interim step is needed; if candidates have made an error and tried to correct it, corrections must be complete to award this A mark

Total: [3]

Question 5(iv):

Answer	Marks	Guidance
$b^2 - 4ac = 0$ seen or used	M1	need not be substituted into; may be stated after formula used or argument towards expressing eqn as a perfect square; e.g. M1 for $(4k-4)^2 - 4 \times 5 \times (k^2 - 16) = 0$
$4k^2 + 32k - 336\ [=0]$ or $k^2 + 8k - 84\ [=0]$	M1	expansion and collection of terms, condoning one error ft their $b^2 - 4ac$; dep on an attempt at $b^2 - 4ac$ with at least two of $a$, $b$ and $c$ correct; may be earned with $< 0$ etc.; may be in formula
use of factorising or quadratic formula or completing the square	M1	condone one error ft; dep on attempt at obtaining required quadratic equation in $k$, not for use with any eqn/inequality they have tried
$k = 6$ or $-14$	A1

or

Answer	Marks	Guidance
Grad of tgt is 2, and normal passes through centre, hence finding equation of normal as $y = -\frac{1}{2}x + 1$ oe	M1
finding $x$ values where diameter $y = -x/2 + 1$ intersects circle as $x = 6$ or $-2$ (condone one error in method)	M1	oe for $y$ values; condone one error in method; or finding intersection of tgt and normal as $\left(\frac{2-2k}{5}, \frac{k+4}{5}\right)$
finding corresponding $y$ values on circle and subst into $y = 2x + k$, or subst their $x$ values into $5x^2 + (4k-4)x + k^2 - 16 = 0$	M1	intersections are $(6, -2)$ and $(-2, 2)$, M0 for just $(6, 2)$ and $(-2, -2)$ used but condone used as well as correct intersections; or subst their intersection of tgt and normal into eqn of circle: $\left(\frac{2-2k}{5}-2\right)^2 + \left(\frac{k+4}{5}\right)^2 = 20$ or ft
$k = 6$ or $-14$	A1	and no other values

Total: [4]

## Question 5(i):

$\sqrt{20}$ isw or $2\sqrt{5}$ | B1 | 0 for $\pm\sqrt{20}$

$(2, 0)$ | B1 |

**Total: [2]**

---

## Question 5(ii):

subst of $x = 0$ into circle eqn soi | M1 | or Pythagoras used on sketch of circle: $2^2 + y^2 = 20$ oe; M0 for just $y^2 = 20$; M1 for $y^2 = 16$ or for $y = 4$; ignore intersections with $x$-axis also found

$y = \pm 4$ oe | A1 | or B2 for just $y = \pm 4$ seen oe; accept both 4 and $-4$ shown on $y$ axis on sketch if both values not stated

sketch of circle with centre $(2, 0)$ or ft their centre from (i) | B1 | if the centre is not marked, it should look roughly correct by eye – coords need not be given on sketch; condone intersections with axes not marked; circle should intersect both $+$ve and $-$ve $x$- and $y$-axes; must be clear attempt at circle; ignore any tangents drawn

**Total: [3]**

---

## Question 5(iii):

$(x-2)^2 + (2x+k)^2 = 20$ | M1 | for attempt to subst $2x + k$ for $y$; allow for attempt to subst $k = y - 2x$ into given eqn

$x^2 - 4x + 4 + 4x^2 + 4kx + k^2 = 20$ | M1 dep | for correct expansion of at least one set of brackets, dependent on first M1; similarly for those working backwards

$5x^2 + (4k-4)x + k^2 - 16 = 0$ | A1 | correct completion to given answer; dependent on both Ms; condone omission of further interim step if both sets of brackets expanded correctly, but for candidates working backwards, at least one interim step is needed; if candidates have made an error and tried to correct it, corrections must be complete to award this A mark

**Total: [3]**

---

## Question 5(iv):

$b^2 - 4ac = 0$ seen or used | M1 | need not be substituted into; may be stated after formula used or argument towards expressing eqn as a perfect square; e.g. M1 for $(4k-4)^2 - 4 \times 5 \times (k^2 - 16) = 0$

$4k^2 + 32k - 336\ [=0]$ or $k^2 + 8k - 84\ [=0]$ | M1 | expansion and collection of terms, condoning one error ft their $b^2 - 4ac$; dep on an attempt at $b^2 - 4ac$ with at least two of $a$, $b$ and $c$ correct; may be earned with $< 0$ etc.; may be in formula

use of factorising or quadratic formula or completing the square | M1 | condone one error ft; dep on attempt at obtaining required quadratic equation in $k$, not for use with any eqn/inequality they have tried

$k = 6$ or $-14$ | A1 |
**or**
Grad of tgt is 2, and normal passes through centre, hence finding equation of normal as $y = -\frac{1}{2}x + 1$ oe | M1 |

finding $x$ values where diameter $y = -x/2 + 1$ intersects circle as $x = 6$ or $-2$ (condone one error in method) | M1 | oe for $y$ values; condone one error in method; or finding intersection of tgt and normal as $\left(\frac{2-2k}{5}, \frac{k+4}{5}\right)$

finding corresponding $y$ values on circle and subst into $y = 2x + k$, or subst their $x$ values into $5x^2 + (4k-4)x + k^2 - 16 = 0$ | M1 | intersections are $(6, -2)$ and $(-2, 2)$, M0 for just $(6, 2)$ and $(-2, -2)$ used but condone used as well as correct intersections; or subst their intersection of tgt and normal into eqn of circle: $\left(\frac{2-2k}{5}-2\right)^2 + \left(\frac{k+4}{5}\right)^2 = 20$ or ft

$k = 6$ or $-14$ | A1 | and no other values

**Total: [4]**

Show LaTeX source

5 A circle has equation $( x - 2 ) ^ { 2 } + y ^ { 2 } = 20$.\\
(i) Write down the radius of the circle and the coordinates of its centre.\\
(ii) Find the points of intersection of the circle with the $y$-axis and sketch the circle.\\
(iii) Show that, where the line $y = 2 x + k$ intersects the circle,

$$5 x ^ { 2 } + ( 4 k - 4 ) x + k ^ { 2 } - 16 = 0$$

(iv) Hence find the values of $k$ for which the line $y = 2 x + k$ is a tangent to the circle.

\hfill \mbox{\textit{OCR MEI C1  Q5 [12]}}

This paper (4 questions)

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Q2 11 Q3 12 Q4 3 Q5 12

Answer	Marks	Guidance
\(\sqrt{20}\) isw or \(2\sqrt{5}\)	B1	0 for \(\pm\sqrt{20}\)
\((2, 0)\)	B1

Answer	Marks	Guidance
subst of \(x = 0\) into circle eqn soi	M1	or Pythagoras used on sketch of circle: \(2^2 + y^2 = 20\) oe; M0 for just \(y^2 = 20\); M1 for \(y^2 = 16\) or for \(y = 4\); ignore intersections with \(x\)-axis also found
\(y = \pm 4\) oe	A1	or B2 for just \(y = \pm 4\) seen oe; accept both 4 and \(-4\) shown on \(y\) axis on sketch if both values not stated
sketch of circle with centre \((2, 0)\) or ft their centre from (i)	B1	if the centre is not marked, it should look roughly correct by eye – coords need not be given on sketch; condone intersections with axes not marked; circle should intersect both \(+\)ve and \(-\)ve \(x\)- and \(y\)-axes; must be clear attempt at circle; ignore any tangents drawn

Answer	Marks	Guidance
\((x-2)^2 + (2x+k)^2 = 20\)	M1	for attempt to subst \(2x + k\) for \(y\); allow for attempt to subst \(k = y - 2x\) into given eqn
\(x^2 - 4x + 4 + 4x^2 + 4kx + k^2 = 20\)	M1 dep	for correct expansion of at least one set of brackets, dependent on first M1; similarly for those working backwards
\(5x^2 + (4k-4)x + k^2 - 16 = 0\)	A1	correct completion to given answer; dependent on both Ms; condone omission of further interim step if both sets of brackets expanded correctly, but for candidates working backwards, at least one interim step is needed; if candidates have made an error and tried to correct it, corrections must be complete to award this A mark

Answer	Marks	Guidance
\(b^2 - 4ac = 0\) seen or used	M1	need not be substituted into; may be stated after formula used or argument towards expressing eqn as a perfect square; e.g. M1 for \((4k-4)^2 - 4 \times 5 \times (k^2 - 16) = 0\)
\(4k^2 + 32k - 336\ [=0]\) or \(k^2 + 8k - 84\ [=0]\)	M1	expansion and collection of terms, condoning one error ft their \(b^2 - 4ac\); dep on an attempt at \(b^2 - 4ac\) with at least two of \(a\), \(b\) and \(c\) correct; may be earned with \(< 0\) etc.; may be in formula
use of factorising or quadratic formula or completing the square	M1	condone one error ft; dep on attempt at obtaining required quadratic equation in \(k\), not for use with any eqn/inequality they have tried
\(k = 6\) or \(-14\)	A1

Answer	Marks	Guidance
Grad of tgt is 2, and normal passes through centre, hence finding equation of normal as \(y = -\frac{1}{2}x + 1\) oe	M1
finding \(x\) values where diameter \(y = -x/2 + 1\) intersects circle as \(x = 6\) or \(-2\) (condone one error in method)	M1	oe for \(y\) values; condone one error in method; or finding intersection of tgt and normal as \(\left(\frac{2-2k}{5}, \frac{k+4}{5}\right)\)
finding corresponding \(y\) values on circle and subst into \(y = 2x + k\), or subst their \(x\) values into \(5x^2 + (4k-4)x + k^2 - 16 = 0\)	M1	intersections are \((6, -2)\) and \((-2, 2)\), M0 for just \((6, 2)\) and \((-2, -2)\) used but condone used as well as correct intersections; or subst their intersection of tgt and normal into eqn of circle: \(\left(\frac{2-2k}{5}-2\right)^2 + \left(\frac{k+4}{5}\right)^2 = 20\) or ft
\(k = 6\) or \(-14\)	A1	and no other values