Coordinate geometry & circles – A-level Maths exam questions with mark scheme

1CoreCalculator3 marks

The points $A$ and $B$ have coordinates $(-1,7)$ and $(5,-2)$ respectively.

Find an equation of the straight line through $A$ and $B$, giving your answer in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers. [3]

for gradient $=\dfrac{-2-7}{5-(-1)}=-\tfrac32$M1
for using $y-y_1=m(x-x_1)$ with their gradient and either point, e.g. $y-7=-\tfrac32(x+1)$M1
$3x+2y-11=0$ (or any integer multiple)A1
Answer: $3x+2y-11=0$
Examiner tip: The final A1 needs integer coefficients and '$=0$' – stopping at $y=-\tfrac32x+\tfrac{11}{2}$ loses it.

2CoreCalculator4 marks

The circle $C$ has equation $$x^{2}+y^{2}-6x+10y-15=0.$$

Find (a) the coordinates of the centre of $C$, [2]

(b) the radius of $C$. [2]

(a) for completing the square in $x$ or $y$: $(x\pm3)^{2}$ or $(y\pm5)^{2}$M1
(a) centre $(3,-5)$A1
(b) for $r^{2}=15+3^{2}+5^{2}$ (their values)M1
(b) $r=7$A1
Answer: (a) $(3,-5)$ (b) $7$
Examiner tip: Watch the signs: $+10y$ gives $(y+5)^{2}$, so the centre has $y=-5$; and add the 15 when finding $r^{2}$, don't subtract it.

3StandardCalculator6 marks

The circle $C$ has centre $(2,-1)$ and passes through the point $P(5,3)$.

(a) Find an equation for $C$. [2]

(b) Find an equation of the tangent to $C$ at $P$, giving your answer in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers. [4]

(a) for $r^{2}=(5-2)^{2}+(3+1)^{2}$M1
(a) $(x-2)^{2}+(y+1)^{2}=25$ (or expanded equivalent $x^{2}+y^{2}-4x+2y-20=0$)A1
(b) gradient of radius $=\dfrac{3-(-1)}{5-2}=\tfrac43$B1
(b) for using perpendicular gradient $-\dfrac{1}{\text{their }m}$M1
(b) for a line through $(5,3)$ with their tangent gradient, e.g. $y-3=-\tfrac34(x-5)$M1
(b) $3x+4y-27=0$ (accept any integer multiple, e.g. $-3x-4y+27=0$ or $6x+8y-54=0$)A1
Answer: (a) $(x-2)^{2}+(y+1)^{2}=25$ (b) $3x+4y-27=0$
Examiner tip: The tangent is perpendicular to the radius at $P$ – use the gradient of $CP$, not of any other line.

4HardCalculator6 marks

The line $l$ has equation $y=x+k$, where $k$ is a constant. The circle $C$ has equation $x^{2}+y^{2}=8$.

(a) Find the values of $k$ for which $l$ is a tangent to $C$. [4]

(b) Hence find the set of values of $k$ for which $l$ does not meet $C$. [2]

(a) for substituting $y=x+k$ into the circle equationM1
(a) $2x^{2}+2kx+k^{2}-8=0$A1
(a) for using $b^{2}-4ac=0$: $4k^{2}-8(k^{2}-8)=0$ (alternative: perpendicular distance from $O$ to $l$, $\dfrac{|k|}{\sqrt2}=\sqrt8$, scores M1 A1 M1 for a complete method)M1
(a) $k=\pm4$A1
(b) for using $b^{2}-4ac<0$ and choosing the outside region of their critical valuesM1
(b) $k<-4$ or $k>4$ (accept $|k|>4$ or set notation; $k<-4$ and $k>4$ scores A0)A1
Answer: (a) $k=4,\ k=-4$ (b) $\{k:k<-4\}\cup\{k:k>4\}$
Examiner tip: 'Does not meet' means $b^{2}-4ac<0$; since $64-4k^{2}<0$ you need the outside region – check with $k=0$, which clearly meets the circle.

5HardCalculator6 marks

The points $A(-3,2)$, $B(5,8)$ and $C(4,1)$ lie on a circle.

(a) Show that angle $ACB=90^\circ$. [3]

(b) Hence find an equation of the circle. [3]

(a) for finding the gradients of $AC$ and $BC$ (or the lengths of all three sides)M1
(a) $m_{AC}=-\tfrac17$ and $m_{BC}=7$ (or $AC^{2}=50,\ BC^{2}=50,\ AB^{2}=100$)A1
(a) $-\tfrac17\times7=-1$ so $AC\perp BC$ (or $50+50=100$, Pythagoras), angle $ACB=90^\circ$A1
(b) recognising $AB$ is a diameter (angle in a semicircle) so the centre is the midpoint $(1,5)$B1
(b) for finding the radius or $r^{2}$ using their centre and $A$ or $B$, e.g. $4^{2}+3^{2}$M1
(b) $(x-1)^{2}+(y-5)^{2}=25$A1
Answer: (a) gradients $-\tfrac17$ and $7$ multiply to $-1$ (b) $(x-1)^{2}+(y-5)^{2}=25$
Examiner tip: The word 'hence' tells you to use (a): a right angle at $C$ means $AB$ is a diameter (angle in a semicircle).

6StretchCalculator7 marks

The circle $C$ has equation $(x-4)^{2}+(y-3)^{2}=10$ and centre $Q$. The point $T$ has coordinates $(-2,1)$.

(a) Show that $T$ lies outside $C$. [2]

Two tangents are drawn from $T$ to $C$, touching $C$ at the points $A$ and $B$, as shown in the diagram.

(b) Find the exact length of $TA$. [3]

(c) Find the exact area of the quadrilateral $TAQB$, giving your answer in the form $p\sqrt{3}$, where $p$ is an integer. [2]

(a) for finding $QT^{2}=(4+2)^{2}+(3-1)^{2}$ or $QT=\sqrt{40}$M1
(a) $\sqrt{40}>\sqrt{10}$ (or $40>10$) so $T$ is outside $C$A1
(b) uses that $QA\perp TA$ (radius perpendicular to tangent), so triangle $TAQ$ is right-angled at $A$B1
(b) for $TA^{2}=QT^{2}-r^{2}=40-10$M1
(b) $TA=\sqrt{30}$A1
(c) for area $=2\times\tfrac12\times\sqrt{30}\times\sqrt{10}$M1
(c) $10\sqrt3$A1
Answer: (a) $QT=\sqrt{40}>\sqrt{10}$ (b) $\sqrt{30}$ (c) $10\sqrt{3}$
Examiner tip: Spot the right angle between radius and tangent – without it you can't use Pythagoras, and that is where the method marks are.

7StandardCalculator6 marks

The line $l$ has equation $2x+3y=12$ and meets the $x$-axis at $A$ and the $y$-axis at $B$. The point $D$ lies on $l$ such that $OD$ is perpendicular to $l$, where $O$ is the origin.

(a) Find the coordinates of $D$. [4]

(b) Find the exact area of triangle $OAD$. [2]

(a) gradient of $l$ is $-\tfrac23$ (may be implied by use of perpendicular gradient $\tfrac32$)B1
(a) for the equation of $OD$ using the perpendicular gradient $-1/m$ through $O$: $y=\tfrac32x$ (or equivalent, e.g. $3x-2y=0$)M1
(a) for solving $2x+3\left(\tfrac32x\right)=12$ (or equivalent) simultaneously to find one coordinateM1
(a) $D\left(\tfrac{24}{13},\tfrac{36}{13}\right)$ (accept awrt $(1.85,\,2.77)$)A1
(b) for $\tfrac12\times OA\times y_D$ with $OA=6$ (or any complete method, e.g. $\tfrac12\times OD\times AD=\tfrac12\times\tfrac{12}{\sqrt{13}}\times\tfrac{18}{\sqrt{13}}$, or the shoelace formula)M1
(b) $\tfrac{108}{13}$ (accept $8\tfrac{4}{13}$; $8.31$ alone is not exact, A0)A1
Answer: (a) $\left(\tfrac{24}{13},\tfrac{36}{13}\right)$ (b) $\tfrac{108}{13}$
Examiner tip: Use $OA$ as the base so the height is just the $y$-coordinate of $D$ – far quicker than finding lengths $OD$ and $AD$.

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