Question 10 - A-Level Maths

OCR C1 — Question 10 12 marks

Exam Board	OCR
Module	C1 (Core Mathematics 1)
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circles
Type	Perpendicular bisector of chord
Difficulty	Moderate -0.5 This is a structured multi-part question testing standard circle geometry concepts (perpendicular bisector of chord, midpoint, circle equation). Each part follows logically from the previous one with clear scaffolding. While it requires multiple steps, all techniques are routine C1 content with no novel problem-solving required, making it slightly easier than average.
Spec	1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 1.03e Complete the square: find centre and radius of circle

10. \includegraphics[max width=\textwidth, alt={}, center]{af6fdbed-fcab-4db8-9cdf-fd049ce720fd-3_668_787_918_431} The diagram shows the circle $C$ and the straight line $l$.
The centre of $C$ lies on the $x$-axis and $l$ intersects $C$ at the points $A ( 2,4 )$ and $B ( 8 , - 8 )$.

Find the gradient of 1 .
Find the coordinates of the mid-point of $A B$.
Find the coordinates of the centre of $C$.
Show that $C$ has the equation $$x ^ { 2 } + y ^ { 2 } - 18 x + 16 = 0$$

Show mark scheme Show mark scheme source

Question 10:

Part (i):

Answer	Marks
$= \frac{-8-4}{8-2} = -2$	M1 A1

Part (ii):

Answer	Marks
$= \left(\frac{2+8}{2}, \frac{4-8}{2}\right) = (5, -2)$	M1 A1

Part (iii):

Answer	Marks
perp. grad $= \frac{-1}{-2} = \frac{1}{2}$	M1
perp. bisector: $y + 2 = \frac{1}{2}(x-5)$	M1 A1
centre where $y = 0$ $\therefore x = 9 \Rightarrow (9, 0)$	M1 A1

Part (iv):

Answer	Marks	Guidance
radius $= \text{dist.}\ (2,4)\ \text{to}\ (9,0) = \sqrt{49+16} = \sqrt{65}$	B1
$\therefore (x-9)^2 + (y-0)^2 = (\sqrt{65})^2$	M1
$x^2 - 18x + 81 + y^2 = 65$
$x^2 + y^2 - 18x + 16 = 0$	A1	(12)

Total: (72)

# Question 10:

## Part (i):
$= \frac{-8-4}{8-2} = -2$ | M1 A1 |

## Part (ii):
$= \left(\frac{2+8}{2}, \frac{4-8}{2}\right) = (5, -2)$ | M1 A1 |

## Part (iii):
perp. grad $= \frac{-1}{-2} = \frac{1}{2}$ | M1 |
perp. bisector: $y + 2 = \frac{1}{2}(x-5)$ | M1 A1 |
centre where $y = 0$ $\therefore x = 9 \Rightarrow (9, 0)$ | M1 A1 |

## Part (iv):
radius $= \text{dist.}\ (2,4)\ \text{to}\ (9,0) = \sqrt{49+16} = \sqrt{65}$ | B1 |
$\therefore (x-9)^2 + (y-0)^2 = (\sqrt{65})^2$ | M1 |
$x^2 - 18x + 81 + y^2 = 65$ | |
$x^2 + y^2 - 18x + 16 = 0$ | A1 | **(12)**

**Total: (72)**

Show LaTeX source

10.\\
\includegraphics[max width=\textwidth, alt={}, center]{af6fdbed-fcab-4db8-9cdf-fd049ce720fd-3_668_787_918_431}

The diagram shows the circle $C$ and the straight line $l$.\\
The centre of $C$ lies on the $x$-axis and $l$ intersects $C$ at the points $A ( 2,4 )$ and $B ( 8 , - 8 )$.\\
(i) Find the gradient of 1 .\\
(ii) Find the coordinates of the mid-point of $A B$.\\
(iii) Find the coordinates of the centre of $C$.\\
(iv) Show that $C$ has the equation

$$x ^ { 2 } + y ^ { 2 } - 18 x + 16 = 0$$

\hfill \mbox{\textit{OCR C1  Q10 [12]}}

This paper (10 questions)

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Q1 3 Q2 4 Q3 5 Q4 6 Q5 7 Q6 7 Q7 8 Q8 10 Q9 10 Q10 12

Answer	Marks
perp. grad \(= \frac{-1}{-2} = \frac{1}{2}\)	M1
perp. bisector: \(y + 2 = \frac{1}{2}(x-5)\)	M1 A1
centre where \(y = 0\) \(\therefore x = 9 \Rightarrow (9, 0)\)	M1 A1

Answer	Marks	Guidance
radius \(= \text{dist.}\ (2,4)\ \text{to}\ (9,0) = \sqrt{49+16} = \sqrt{65}\)	B1
\(\therefore (x-9)^2 + (y-0)^2 = (\sqrt{65})^2\)	M1
\(x^2 - 18x + 81 + y^2 = 65\)
\(x^2 + y^2 - 18x + 16 = 0\)	A1	(12)