Mathematics 9709 · AS & A Level
Coordinate geometry
100 practice questions on Coordinate geometry, with worked solutions and instant marking.
The diagram shows a section of the curve $y = \frac{1}{16}(3x - 1)^2$, which is tangent to the $x$-axis at $P$. The point $Q(3, 4)$ lies on the curve, and the tangent at $Q$ meets the $x$-axis at $R$.
Feb/March 2016
The coordinates of two points are $A(5, 7)$ and $B(9, -1)$.
Feb/March 2016
The diagram depicts a pyramid $OABC$ with a flat triangular base $OAB$ and perpendicular height $OC$. Angles $AOB$, $BOC$ and $AOC$ are each right angles. Unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ run parallel to $OA$, $OB$ and $OC$ respectively, and $OA = 4$ units, $OB = 2.4$ units and $OC = 3$ units. Point $P$ lies on $CA$ so that $CP = 3$ units.
Feb/March 2016
The diagram displays the graphs of $y = \tan x$ and $y = \cos x$ for $0 \leq x \leq \pi$. They meet at points $A$ and $B$.
Feb/March 2017
Taking origin $O$ as the reference point, the position vectors of $A$ and $B$ are $overrightarrow{OA} = 2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}$ and $\overrightarrow{OB} = 7\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$.
Feb/March 2017
The vectors $\mathbf{u}$ and $\mathbf{v}$ are given by $\mathbf{u} = \begin{pmatrix} q \\ 2 \\ 6 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 8 \\ q-1 \\ q^2-7 \end{pmatrix}$, with $q$ as a constant.
Feb/March 2019
The endpoints of a diameter of circle $C_1$ are $(-3, -5)$ and $(7, 3)$. Circle $C_1$ is shifted by $\begin{pmatrix}8\\4\end{pmatrix}$ to form circle $C_2$, as the diagram shows.
Feb/March 2020
The curve is given by $y = x^2 - 2x - 3$. A point travels along the curve so that, at $P$, the $y$-coordinate is increasing at 4 units per second and the $x$-coordinate is increasing at 6 units per second.
Feb/March 2020
A line is given by the equation $y = 3x + k$ and a curve is given by the equation $y = x^2 + kx + 6$, with $k$ a constant.
Feb/March 2021
The points $A(7, 1)$, $B(7, 9)$ and $C(1, 9)$ lie on a circle’s circumference.
Feb/March 2021
In the diagram, the circle has equation $(x + 1)^2 + (y - 2)^2 = 85$ and the straight line has equation $y = 3x - 20$. The line cuts the circle at $A$ and $B$, and the centre of the circle is $C$.
Feb/March 2022
The equation of a line is $y = 3x - 2k$ and the equation of a curve is $y = x^2 - kx + 2$, with $k$ constant.
Feb/March 2023
Points $A\,(7,12)$ and $B$ are on a circle centred at $(-2,5)$. The equation of line $AB$ is $y = -2x + 26$.
Feb/March 2023
The circle has centre $C(-4, 5)$ and radius $\sqrt{20}$ units, as shown in the diagram. It meets the $y$-axis at $A$ and $B$. The angle $ACB$ has size $\theta$ radians.
Feb/March 2024
The line $y = x + 5$ intersects the curve $2x^2 + 3y^2 = k$ at one only point, $P$.
Feb/March 2024
The diagram depicts triangle $OAB$ with $OA = OB = 10\text{ cm}$ and $\angle AOB = 0.8$ radians. Let $C$ and $D$ lie on $OA$ and $OB$ respectively, so that the arc $CD$ belongs to a circle centred at $O$ with radius $6\text{ cm}$. The shaded part is enclosed by the arc $CD$ together with the line segments $CA$, $AB$ and $BD$.
Feb/March 2025
The figure depicts a circle $C$ with radius $r$, and every point on $C$ has $x > 0$ and $y > 0$. The smallest distance from any point on $C$ to the $x$-axis is $8$ units, while the smallest distance from any point on $C$ to the $y$-axis is $5$ units.
Feb/March 2025
The diagram depicts parallelogram $OABC$. You are given that $\vec{OA} = i + 3j + 3k$ and $\vec{OC} = 3i - j + k$.
May/June 2010
The diagram depicts a section of the curve $y = 2 - \frac{18}{2x + 3}$, which meets the $x$-axis at $A$ and the $y$-axis at $B$. The normal drawn to the curve at $A$ meets the $y$-axis at $C$.
May/June 2010
The diagram presents triangle $ABC$, with $A$ at $(3, -2)$ and $B$ at $(15, 22)$. The gradients of $AB$, $AC$ and $BC$ are $2m$, $-2m$ and $m$ respectively, where $m$ is a positive constant.
May/June 2010
From the diagram, $A$ has the coordinates $(-1,3)$ and $B$ has the coordinates $(3,1)$. Line $L_1$ goes through $A$ and is parallel to $OB$. Line $L_2$ goes through $B$ and is perpendicular to $AB$. The lines $L_1$ and $L_2$ intersect at $C$.
May/June 2010
With respect to an origin $O$, the position vectors of the points $A$ and $B$ are $\vec{OA} = \begin{pmatrix}-2 \\ 3 \\ 1\end{pmatrix}$ and $\vec{OB} = \begin{pmatrix}4 \\ 1 \\ p\end{pmatrix}$.
May/June 2010
With respect to an origin $O$, the position vectors of points $A$, $B$ and $C$ are $overrightarrow{OA} = \mathbf{i} - 2\mathbf{j} + 4\mathbf{k}$, $\overrightarrow{OB} = 3\mathbf{i} + 2\mathbf{j} + 8\mathbf{k}$, and $\overrightarrow{OC} = -\mathbf{i} - 2\mathbf{j} + 10\mathbf{k}$.
May/June 2010
The diagram depicts a rhombus $ABCD$ with $A$ at $(-1, 2)$, $C$ at $(5, 4)$, and $B$ located on the $y$-axis.
May/June 2010
The diagram depicts a prism $ABCDPQRS$ with a horizontal square base $APSD$ of side length $6\text{ cm}$. The cross-section $ABCD$ is a trapezium, and the vertical edges $AB$ and $DC$ measure $5\text{ cm}$ and $2\text{ cm}$ respectively. Unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are parallel to $AD$, $AP$ and $AB$ respectively.
May/June 2011
Line $L_1$ goes through points $A\,(2,5)$ and $B\,(10,9)$. Line $L_2$ runs parallel to $L_1$ and passes through the origin. Point $C$ is on $L_2$ in such a way that $AC$ is perpendicular to $L_2$.
May/June 2011
With respect to the origin $O$, the position vectors for the points $A$, $B$ and $C$ are $\overrightarrow{OA} = \begin{pmatrix}2\\3\\5\end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix}4\\2\\3\end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix}10\\0\\6\end{pmatrix}$.
May/June 2011
The line $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ are positive constants, cuts the $x$-axis at $P$ and the $y$-axis at $Q$. Given that $PQ = \sqrt{45}$ and that the gradient of the line $PQ$ is $-\frac{1}{2}$, determine the values of $a$ and $b$.
May/June 2011
The diagram shows $OABCDEFG$ as a rectangular block with $OA = OD = 6$ cm and $AB = 12$ cm. The unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ run parallel to $OA$, $OC$ and $OD$ respectively. Point $P$ is the midpoint of $DG$, $Q$ is at the centre of the square face $CBFG$, and $R$ is located on $AB$ so that $AR = 4$ cm.
May/June 2011
The vectors $\mathbf{u}$ and $\mathbf{v}$ are given by $\mathbf{u} = \begin{pmatrix} p^2 \\ -2 \\ 6 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 2 \\ p - 1 \\ 2p + 1 \end{pmatrix}$, with $p$ as a constant.
May/June 2012
Point $A$ has coordinates $(-3, 2)$, while point $C$ has coordinates $(5, 6)$. The midpoint of $AC$ is $M$, and the perpendicular bisector of $AC$ meets the $x$-axis at $B$.
May/June 2012
Point $A$ is at $(-1, -5)$ and point $B$ is at $(7, 1)$. The perpendicular bisector of $AB$ crosses the $x$-axis at $C$ and the $y$-axis at $D$. Find the length of $CD$.
May/June 2012
The diagram depicts a metal plate formed by cutting out a segment from a circle with centre $O$ and radius $8$ cm. The line $AB$ is a chord of the circle, and the angle $AOB = 2.4$ radians.
May/June 2012
Find the angle between the vectors $3\mathbf{i} - 4\mathbf{k}$ and $2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$.
May/June 2012
The line is given by the equation $2y + x = k$, where $k$ is constant, and the curve is described by $xy = 6$.
May/June 2012
With respect to origin $O$, the position vectors of points $A$, $B$ and $C$ are $\overrightarrow{OA} = \begin{pmatrix}2 \\ -1 \\ 4\end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix}4 \\ 2 \\ -2\end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix}1 \\ 3 \\ p\end{pmatrix}$.
May/June 2012
The curve $y = \frac{10}{2x + 1} - 2$ cuts the $x$-axis at $A$, and the tangent at $A$ meets the $y$-axis at $C$.
May/June 2012
The diagram includes a section of the curve $y = (x - 2)^4$ and the point $A(1, 1)$ on that curve. The tangent drawn at $A$ meets the $x$-axis at $B$, while the normal drawn at $A$ meets the $y$-axis at $C$.
May/June 2013
Taking $O$ as the origin, the position vectors of the three points $A$, $B$ and $C$ are $\vec{OA} = \mathbf{i} + 2p\mathbf{j} + q\mathbf{k}$, $\vec{OB} = q\mathbf{i} - 2p\mathbf{k}$ and $\vec{OC} = -(4p^2 + q^2)\mathbf{i} + 2p\mathbf{j} + q\mathbf{k}$, where $p$ and $q$ are constants.
May/June 2013
The sketch displays the curve $y = \sqrt{1 + 4x}$, which cuts the $x$-axis at $A$ and the $y$-axis at $B$. The normal drawn to the curve at $B$ meets the $x$-axis at $C$.
May/June 2013
At point $P$, the straight line $y = mx + 14$ touches the curve $y = \frac{12}{x} + 2$.
May/June 2013
Taking O as the origin, the position vectors of points A and B are given as \(\overrightarrow{OA} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\) and \(\overrightarrow{OB} = 3\mathbf{i} + p\mathbf{j} + q\mathbf{k}\), with p and q as constants.
May/June 2013
The point $R$ is the image of $(-1, 3)$ after reflection in the line $3y + 2x = 33$.
May/June 2013
The diagram presents a segment of the curve $y = \frac{8}{\sqrt{x}} - x$ together with the points $A(1,7)$ and $B(4,0)$, both of which are on the curve. The tangent to the curve at $B$ meets the line $x = 1$ at $C$.
May/June 2013
The diagram illustrates three points $A(2,14)$, $B(14,6)$ and $C(7,2)$. Point $X$ is located on $AB$, and $CX$ is at right angles to $AB$.
May/June 2013
The diagram depicts a parallelogram $OABC$ in which $\vec{OA} = \begin{pmatrix}3\\3\\-4\end{pmatrix}$ and $\vec{OB} = \begin{pmatrix}5\\0\\2\end{pmatrix}$.
May/June 2013
The points $A$ and $B$ have coordinates $(a, 2)$ and $(3, b)$ respectively, where $a$ and $b$ are constants. The distance $AB$ is $\sqrt{125}$ units and the gradient of the line $AB$ is $2$. Determine the possible values of $a$ and $b$.
May/June 2014
Using origin $O$ as the reference point, the position vectors for points $A$ and $B$ are $\overrightarrow{OA} = \begin{pmatrix}3p\\4\\p^2\end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix}-p\\-1\\p^2\end{pmatrix}$.
May/June 2014
Determine the coordinates of the point where the perpendicular bisector of the line joining $(2, 7)$ to $(10, 3)$ crosses the $x$-axis.
May/June 2014
The diagram represents trapezium $ABCD$, with $BA$ parallel to $CD$. Relative to an origin $O$, the position vectors of $A$, $B$ and $C$ are given by $\overrightarrow{OA} = \begin{pmatrix}3\\4\\0\end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix}1\\3\\2\end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix}4\\5\\6\end{pmatrix}$.
May/June 2014
The diagram illustrates a parallelogram $ABCD$, in which the equation for $AB$ is $y = 3x$ and the equation for $AD$ is $4y = x + 11$. The diagonals $AC$ and $BD$ intersect at the point $E\left(6\frac{1}{2}, 8\frac{1}{2}\right)$.
May/June 2014
Measured from the origin $O$, the position vectors of points $A$ and $B$ are $ \overrightarrow{OA} = \begin{pmatrix}3 \\ 0 \\ -4\end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix}6 \\ -3 \\ 2\end{pmatrix}$. The position vector of $C$ is $\overrightarrow{OC} = \begin{pmatrix}k \\ -2k \\ 2k - 3\end{pmatrix}$.
May/June 2015
The straight line with gradient $-2$ that passes through $P(3t, 2t)$ cuts the $x$-axis at $A$ and the $y$-axis at $B$. The line through $P$ that is perpendicular to $AB$ meets the $x$-axis at $C$.
May/June 2015
The point $C$ is located on the perpendicular bisector of the segment joining $A(4, 6)$ and $B(10, 2)$. $C$ is also on the line through $(3, 11)$ that runs parallel to $AB$.
May/June 2015
With $O$ taken as the origin, the position vectors of $A$ and $B$ are $\overrightarrow{OA} = 2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}$ and $\overrightarrow{OB} = 3\mathbf{i} + \mathbf{j} + 4\mathbf{k}$. Point $C$ is defined so that $\overrightarrow{AB} = \overrightarrow{BC}$.
May/June 2015
Taking $O$ as the origin, the position vectors of the points $A$, $B$ and $C$ are $overrightarrow{OA} = \begin{pmatrix} 3 \\ 2 \\ -3 \end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix} 6 \\ 1 \\ 2 \end{pmatrix}$.
May/June 2015
Point $A$ is at $(p, 1)$, while point $B$ is at $(9, 3p + 1)$, with $p$ fixed as a constant.
May/June 2015
With origin $O$, the position vectors of $A$, $B$ and $C$ are $\vec{OA} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$, $\vec{OB} = \begin{pmatrix} 5 \\ -1 \\ k \end{pmatrix}$ and $\vec{OC} = \begin{pmatrix} 2 \\ 6 \\ -3 \end{pmatrix}$ respectively, where $k$ is a constant.
May/June 2016
With origin $O$, the position vectors of $A$ and $B$ are $\overrightarrow{OA} = 2\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}$ and $\overrightarrow{OB} = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}$, respectively.
May/June 2016
Three points are given by the coordinates $A(0, 7)$, $B(8, 3)$ and $C(3k, k)$. Find the value of the constant $k$ for which
May/June 2016
The vertices of triangle $ABC$ are $A(-2, -1)$, $B(4, 6)$ and $C(6, -3)$.
May/June 2016
The position vectors of $A$, $B$ and $C$ from the origin $O$ are $\vec{OA} = \begin{pmatrix}2\\3\\-4\end{pmatrix}$, $\vec{OB} = \begin{pmatrix}1\\5\\p\end{pmatrix}$ and $\vec{OC} = \begin{pmatrix}5\\0\\2\end{pmatrix}$, with $p$ as a constant.
May/June 2016
Taking origin $O$, the position vectors of points $A$ and $B$ are given by $\overrightarrow{OA} = \begin{pmatrix} 3 \\ -6 \\ p \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 2 \\ -6 \\ -7 \end{pmatrix}$, and angle $AOB = 90^\circ$. Point $C$ is defined so that $\overrightarrow{OC} = \frac{2}{3}\overrightarrow{OA}$.
May/June 2017
Point $A$ is at $(-2, 6)$. The perpendicular bisector of line $AB$ is given by $2y = 3x + 5$.
May/June 2017
Using origin $O$, the position vectors of the points $A$, $B$ and $C$ are $ \overrightarrow{OA} = 3\mathbf{i} + p\mathbf{j} - 2p\mathbf{k}$, $\overrightarrow{OB} = 6\mathbf{i} + (p + 4)\mathbf{j} + 3\mathbf{k}$ and $\overrightarrow{OC} = (p - 1)\mathbf{i} + 2\mathbf{j} + q\mathbf{k}$, with $p$ and $q$ as constants.
May/June 2017
Taking $O$ as the origin, the position vectors of $A$ and $B$ are $\overrightarrow{OA} = \begin{pmatrix}5\\1\\-2\end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix}5\\4\\-3\end{pmatrix}$. The point $P$ is on $AB$ and satisfies $\overrightarrow{AP} = \frac{1}{3}\overrightarrow{AB}$.
May/June 2017
The points $A(-1, 1)$ and $P(a, b)$ are given, where $a$ and $b$ are constants. The gradient of $AP$ is $2$.
May/June 2017
A curve is defined by $\frac{dy}{dx} = \frac{12}{(2x + 1)^2}$. The point $(1, 1)$ is on the curve.
May/June 2018
The diagram depicts a kite $OABC$, with $AC$ as the axis of symmetry. The coordinates of $A$ and $C$ are $(0,4)$ and $(8,0)$, respectively, and $O$ is the origin.
May/June 2018
Relative to origin $O$, the position vectors of points $A$, $B$ and $C$ are $\overrightarrow{OA} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$.
May/June 2018
The diagram represents a three-dimensional shape. The base $OAB$ is a horizontal triangle with angle $AOB$ equal to $90^\circ$. The side $OBCD$ is a rectangle, and the side $OAD$ lies in a vertical plane. Unit vectors $\mathbf{i}$ and $\mathbf{j}$ are parallel to $OA$ and $OB$ respectively, and the unit vector $\mathbf{k}$ is vertical. The position vectors of $A$, $B$ and $D$ are given by $\overrightarrow{OA} = 8\mathbf{i}$, $\overrightarrow{OB} = 5\mathbf{j}$ and $\overrightarrow{OD} = 2\mathbf{i} + 4\mathbf{k}$.
May/June 2018
The points A and B are located at (h, h) and (4h + 6, 5h) respectively. The perpendicular bisector of AB is given by 3x + 2y = k.
May/June 2018
Points $A$ and $B$ have coordinates $(-3k - 1,\ k + 3)$ and $(k + 3,\ 3k + 5)$ respectively, where $k$ is a constant $(k \neq -1)$.
May/June 2018
The diagram depicts a pyramid $OABCD$ with a horizontal rectangular base $OABC$. The edges $OA$ and $AB$ are $8$ units and $6$ units long respectively. Point $E$ lies on $OB$ in such a way that $OE = 2$ units. The vertex $D$ of the pyramid is $7$ units vertically above $E$. Unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are parallel to $OA$, $OC$ and $OE$ respectively.
May/June 2018
At the point $P$ on the curve, the line $4y = x + c$, with $c$ constant, touches the curve $y^2 = x + 3$ as a tangent.
May/June 2019
The diagram depicts a trapezium $ABCD$ with $A$, $B$ and $C$ at $(4, 0)$, $(0, 2)$ and $(h, 3h)$ respectively. $BC$ is parallel to $AD$, $ngle ABC = 90^\circ$ and $CD$ is parallel to the $x$-axis.
May/June 2019
The sketch represents a three-dimensional solid whose base $OABC$ and top face $DEFG$ are matching horizontal squares. The parallelograms $OAED$ and $CBFG$ each lie in vertical planes. Point $M$ is the midpoint of $AF$. The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are parallel to $OA$ and $OC$ respectively, while $\mathbf{k}$ points vertically upwards. The position vectors of $A$ and $D$ are given by $\overrightarrow{OA} = 8\mathbf{i}$ and $\overrightarrow{OD} = 3\mathbf{i} + 10\mathbf{k}$.
May/June 2019
The coordinates of the points $A$ and $B$ are $(1, 3)$ and $(9, -1)$ respectively. The perpendicular bisector of $AB$ cuts the $y$-axis at point $C$. Determine the coordinates of $C$.
May/June 2019
Relative to the origin $O$, the position vectors of points $A$ and $B$ are $\vec{OA} = \begin{pmatrix} 6 \\ -2 \\ -6 \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} 3 \\ k \\ -3 \end{pmatrix}$, with $k$ a constant.
May/June 2019
The figure depicts a solid $ABCDEF$ whose horizontal base $ABC$ is a triangle with a right angle at $A$. $AB$ and $AC$ are $8$ units and $6$ units long respectively, and $M$ is the midpoint of $AB$. Point $D$ is $7$ units directly above $A$. Triangle $DEF$ is in a horizontal plane, with $DE$, $DF$ and $FE$ parallel to $AB$, $AC$ and $CB$ respectively, and $N$ is the midpoint of $FE$. The lengths of $DE$ and $DF$ are $4$ units and $2$ units respectively. Unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are parallel to $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ respectively.
May/June 2019
Points $A$ and $B$ have coordinates $(1,3)$ and $(9,-1)$ respectively, and $D$ is the mid-point of $AB$. Point $C$ is located at $(x,y)$, where $x$ and $y$ are variables.
May/June 2019
The coordinates of points $A$ and $B$ are $(-1,-2)$ and $(7,4)$, respectively.
May/June 2020
The diagram shows $ABC$ as a semicircle with diameter $AC$, centre $O$ and radius $6\text{ cm}$. Arc $AB$ measures $15\text{ cm}$. Point $X$ is on $AC$ and $BX$ is perpendicular to $AX$.
May/June 2020
For the circle with centre $C$, the equation is $x^2 + y^2 - 8x + 4y - 5 = 0$.
May/June 2020
Points A and B have coordinates $(-7, 3)$ and $(5, 11)$, respectively.
May/June 2020
The circle is given by the equation $x^2 + y^2 - 4x + 6y - 77 = 0$.
May/June 2021
The coordinates of points $A$ and $B$ are $(8, 3)$ and $(p, q)$ respectively. The perpendicular bisector of $AB$ is given by $y = -2x + 4$.
May/June 2021
Point $A$ is located at $(1, 5)$, and line $l$ passes through $A$ with gradient $-\frac{2}{3}$. The circle is centred at $(5, 11)$ and has radius $\sqrt{52}$.
May/June 2021
The points $A(-2, 3)$, $B(3, 0)$ and $C(6, 5)$ are positioned on the circumference of a circle whose centre is $D$.
May/June 2021
The circle is given by the equation $x^2 + y^2 + 6x - 2y - 26 = 0$.
May/June 2022
The circle is given by the equation $x^2 + y^2 + ax + by - 12 = 0$. The points $A(1, 1)$ and $B(2, -6)$ are located on the circle.
May/June 2022
The diagram illustrates the circle whose equation is $(x - 2)^2 + (y + 4)^2 = 20$, with centre $C$. Point $B$ is at $(0, 2)$, and the line segment $BC$ cuts the circle at $P$.
May/June 2022
The diagram displays circle $P$, whose centre is $(0, 2)$ and whose radius is $10$, together with the tangent to the circle at $A(6, 10)$. It also displays a second circle $Q$, centred at the point where this tangent crosses the $y$-axis, with radius $\frac{5}{2}\sqrt{5}$.
May/June 2023
The diagram depicts a sector $ABC$ of a circle whose centre is $A$ and whose radius is $8\text{ cm}$. The sector has area $\frac{16}{3}\pi\text{ cm}^2$. Point $D$ is located on arc $BC$.
May/June 2023
A circle has equation $(x-a)^2+(y-3)^2=20$. The straight line $y=\frac{1}{2}x+6$ is tangent to the circle at point $P$.
May/June 2023
The circle is given by $(x - 1)^2 + (y + 4)^2 = 40$, and the line $y = x - 9$ cuts it at $A$ and $B$.
May/June 2023
The diagram depicts sector $OAB$ from a circle with centre $O$ and radius $r$ cm. The angle $AOB = \theta$ radians. You are told that arc length $AB$ is $9.6$ cm and that sector area $OAB$ is $76.8$ cm$^2$.
May/June 2023
The curve, which goes through $(0, 3)$, is described by $y = f(x)$. Also given is $f'(x) = 1 - \frac{2}{(x-1)^3}$.
May/June 2023
The circle is given by $(x - 3)^2 + y^2 = 18$. A line of the form $y = mx + c$ passes through $(0, -9)$ and is tangent to the circle.
May/June 2024
A circle is given by $(x - 6)^2 + (y + a)^2 = 18$. The line $y = 2a - x$ touches the circle, so it is a tangent.
May/June 2024