Analytical geometry — Mathematics Grade 12

6 exam-style question sets on Analytical geometry (Paper 2), each with a hint and a fully worked answer. The app holds 30 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Right-angled triangle: inclination, parallel line, area & circumcentreTwo circles: common tangent & internal contactTriangle: gradients, lengths, ratios & areasCircle: centre, tangents & inscribed anglesTriangle & parallelogram: surd length, inclination, foot of perpendicular & areaCircle: equation, tangents from an external point

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Question 1

Right-angled triangle: inclination, parallel line, area & circumcentreRoutine20 marksPaper 2

In the diagram, $A (3; - 7)$ , $B (- 3; - 4)$ and $C (1; 4)$ are the vertices of $△ A B C$ , with $A \hat{B} C = 9 0^{\circ}$ . Side $B C$ cuts the $x$ -axis at $D$ , and $θ$ is the angle of inclination of $B C$ .

aCalculate the length of $A B$ . Leave your answer in simplest surd form. (2)
bDetermine the gradient of $B C$ . (2)
cCalculate the size of $θ$ , correct to TWO decimal places. (2)
dCalculate the size of $A \hat{C} B$ . (3)
eDetermine the equation of the line through $A$ that is parallel to $B C$ . Give your answer in the form $y = m x + c$ . (3)
fCalculate the area of $△ A B C$ . (3)
g $P$ is the point that is equidistant from $A$ , $B$ and $C$ . Determine the coordinates of $P$ . Give a reason for your answer. (3)
hHence, calculate the size of $A \hat{P} B$ . Give a reason for your answer. (2)

This question has a diagram, shown in the app.

Hint

Use the distance and gradient formulae for the first parts. For $A \hat{C} B$ you need the inclinations of BOTH lines through $C$ : the exterior angle of the triangle that these two lines form with the $x$ -axis links them. For $P$ , ask what the right angle at $B$ tells you about side $A C$ in the circle through $A$ , $B$ and $C$ ; the last part then follows from a circle theorem.

Worked answer

a) $A B = (3 - (- 3))^{2} + (- 7 - (- 4))^{2}$ ✓ $= 45 = 35 \approx 6, 71$ units ✓

b) $m_{B C} = \frac{4 - ( - 4 )}{1 - ( - 3 )} = \frac{8}{4}$ ✓ $= 2$ ✓

c) $tan θ = m_{B C} = 2$ ✓ $\Rightarrow θ = 63, 4 3^{\circ}$ ✓

d) $m_{A C} = \frac{4 - ( - 7 )}{1 - 3} = - \frac{11}{2}$ ✓ Let $β$ be the inclination of $A C$ : $tan β = - \frac{11}{2}$ , and the gradient is negative, so $β = 18 0^{\circ} - 79, 7 0^{\circ} = 100, 3 0^{\circ}$ ✓ In the triangle formed by $C B$ , $C A$ and the $x$ -axis: $β = θ + A \hat{C} B$ (ext ∠ of Δ), so $A \hat{C} B = 100, 3 0^{\circ} - 63, 4 3^{\circ} = 36, 8 7^{\circ}$ ✓

e) The line is parallel to $B C$ , so $m = m_{B C} = 2$ ✓ Substitute $A (3; - 7)$ : $- 7 = 2 (3) + c$ ✓ $\Rightarrow c = - 13$ , so $y = 2 x - 13$ ✓

f) $B C = (1 - (- 3))^{2} + (4 - (- 4))^{2} = 80 = 45$ ✓ Since $A B ⊥ B C$ : area of $△ A B C = \frac{1}{2} \times A B \times B C = \frac{1}{2} \times 35 \times 45$ ✓ $= 30$ square units ✓

g) $A \hat{B} C = 9 0^{\circ}$ , so $A C$ is a diameter of the circle through $A$ , $B$ and $C$ (converse ∠ in semi-circle) ✓ The point equidistant from $A$ , $B$ and $C$ is the centre of this circle, i.e. the midpoint of the diameter $A C$ ✓ $P = (\frac{3 + 1}{2}; \frac{- 7 + 4}{2}) = (2; - \frac{3}{2})$ ✓ (Check: $P A = P B = P C = \frac{5 5}{2}$ .)

h) $P$ is the centre of the circle and $A \hat{C} B$ stands on the same arc $A B$ , so $A \hat{P} B = 2 \times A \hat{C} B$ (∠ at centre $= 2 \times$ ∠ at circumf.) ✓ $= 2 \times 36, 8 7^{\circ} = 73, 7 4^{\circ}$ ✓

Modelled on Nov 2023 Paper 2, Question 3 — same skills, new scenario.

Question 2

Two circles: common tangent & internal contactRoutine20 marksPaper 2

In the diagram, $O$ is the centre of the circle $x^{2} + y^{2} = 52$ . A second, larger circle has its centre $M (n; 0)$ on the $x$ -axis; the two circles do not meet. The straight line $P R$ is a tangent to both circles. It meets the smaller circle only at $P (6; q)$ , a point in the first quadrant, and the larger circle only at $R$ . The circles lie on opposite sides of $P R$ . The point $S (8; 1)$ lies on $P R$ , between $P$ and $R$ , such that $S R = 3 P S$ .

aShow that $q = 4$ . (2)
bDetermine the coordinates of $R$ . (3)
cFind the equation of the tangent $P R$ . Give your answer in the form $y = m x + c$ . (4)
dCalculate the value of $n$ . (3)
eDetermine the equation of the circle centred at $M$ , giving your answer in the form $x^{2} + y^{2} + A x + B y + C = 0$ . (4)
fThe circle centred at $O$ is translated $k$ units horizontally so that it now lies inside the larger circle and touches it at exactly one point. Calculate ALL possible values of $k$ . Leave your answers in simplest surd form. (4)

This question has a diagram, shown in the app.

Hint

Substitute $P$ into $x^{2} + y^{2} = 52$ and let the quadrant decide the sign. Along a straight line the changes in $x$ and in $y$ stay in proportion, so $S R = 3 P S$ means the step from $S$ to $R$ is three times the step from $P$ to $S$ . A tangent is perpendicular to the radius at the point of contact — at $P$ AND at $R$ — so $M R ∥ O P$ . For the last part, circles that touch internally have centres exactly (larger radius minus smaller radius) apart, and the shifted centre may lie on either side of $M$ .

Worked answer

a) $6^{2} + q^{2} = 52$ ✓
$q^{2} = 16$ , so $q = \pm 4$ ; $P$ lies in the first quadrant, so $q = 4$ ✓

b) $P$ , $S$ and $R$ lie on one straight line and $S R = 3 P S$ , so the change in $x$ and in $y$ from $S$ to $R$ is three times the change from $P$ to $S$ ✓
From $P (6; 4)$ to $S (8; 1)$ : $x$ increases by $2$ and $y$ decreases by $3$
$x_{R} = 8 + 3 (2) = 14$ ✓ and $y_{R} = 1 + 3 (- 3) = - 8$ ✓
$∴ R (14; - 8)$

c) $m_{O P} = \frac{4 - 0}{6 - 0} = \frac{2}{3}$ ✓
$P R ⊥ O P$ (tan $⊥$ radius), so $m_{P R} = - \frac{3}{2}$ ✓
$y - 4 = - \frac{3}{2} (x - 6)$ ✓
$y = - \frac{3}{2} x + 13$ ✓
(Check: $- \frac{3}{2} (8) + 13 = 1$ and $- \frac{3}{2} (14) + 13 = - 8$ , so $S$ and $R$ lie on this line.)

d) $M R ⊥ P R$ (tan $⊥$ radius) and $O P ⊥ P R$ , so $M R ∥ O P$ and $m_{M R} = \frac{2}{3}$ ✓
$\frac{0 - ( - 8 )}{n - 14} = \frac{2}{3}$ ✓
$24 = 2 n - 28 \Rightarrow 2 n = 52 \Rightarrow n = 26$ ✓

e) $M R^{2} = (14 - 26)^{2} + (- 8 - 0)^{2}$ ✓ $= 144 + 64 = 208$ ✓
$(x - 26)^{2} + y^{2} = 208$ ✓
$x^{2} - 52 x + 676 + y^{2} - 208 = 0$
$x^{2} + y^{2} - 52 x + 468 = 0$ ✓

f) Radii: $O P = 52 = 213$ and $M R = 208 = 413$ ✓
Internal contact: the distance between the centres $= 413 - 213 = 213$ ✓
The shifted centre is $(k; 0)$ , so $k - 26 = \pm 213$
$k = 26 + 213$ ✓ or $k = 26 - 213$ ✓ (approximately $33, 21$ or $18, 79$ )

Modelled on Nov 2023 Paper 2, Question 4 — same skills, new scenario.

Question 3

Triangle: gradients, lengths, ratios & areasProblem-solving19 marksPaper 2

In the diagram, $△ P QR$ is drawn with $Q (- 4; 5)$ and $R (6; 0)$ . The vertex $P$ lies in the second quadrant. $T (2; t)$ is a point on side $QR$ , and $N$ is a point on side $QP$ such that $T N ∥ R P$ .

aCalculate the gradient of $QR$ . (2)
bDetermine the equation of the line $QR$ in the form $y = m x + c$ . (2)
cShow that $t = 2$ . (1)
dCalculate the length of $QR$ . Leave your answer in simplest surd form. (2)
eDetermine, in its simplest form, the ratio $QT : QR$ . (2)
fDetermine, giving reasons, the ratio of the area of $△ QT N$ to the area of $△ QR P$ . (4)
gThe gradient of $QP$ is $3$ and $QP = 210$ units. Calculate the coordinates of $P$ . (6)

This question has a diagram, shown in the app.

Hint

Work along the line $QR$ for the first five parts; $T N ∥ R P$ makes $△ QT N$ similar to $△ QR P$ , and for $P$ combine the gradient condition with the distance formula before testing the quadrant.

Worked answer

a) $m_{QR} = \frac{0 - 5}{6 - ( - 4 )}$ ✓ $= - \frac{1}{2}$ ✓

b) $y = - \frac{1}{2} x + c$ ✓ Substitute $R (6; 0)$ : $0 = - \frac{1}{2} (6) + c$ , so $c = 3$ ✓ Thus $y = - \frac{1}{2} x + 3$

c) Substitute $x = 2$ into the line $QR$ : $y = - \frac{1}{2} (2) + 3 = 2$ , so $t = 2$ ✓

d) $QR = (6 - (- 4))^{2} + (0 - 5)^{2} = 125$ ✓ $= 55 \approx 11, 18$ units ✓

e) $Q$ , $T$ and $R$ are collinear, so $\frac{QT}{QR} = \frac{x _{T} - x _{Q}}{x _{R} - x _{Q}} = \frac{2 - ( - 4 )}{6 - ( - 4 )} = \frac{6}{10}$ ✓ $∴ QT : QR = 3 : 5$ ✓

f) In $△ QT N$ and $△ QR P$ : $\hat{Q}$ is common and $Q \hat{T} N = Q \hat{R} P$ (corresp $∠$ s; $T N ∥ R P$ ) ✓ $∴ △ QT N ∣∣∣△ QR P$ (equiangular $△$ s) ✓ $\frac{area △ QT N}{area △ QR P} = (\frac{QT}{QR})^{2}$ ✓ (areas of similar $△$ s) $= (\frac{3}{5})^{2} = \frac{9}{25}$ ✓

g) Let $P (x; y)$ . Gradient of $QP$ : $y - 5 = 3 (x + 4)$ ✓ Distance: $(x + 4)^{2} + (y - 5)^{2} = (210)^{2} = 40$ ✓ Substitute: $(x + 4)^{2} + 9 (x + 4)^{2} = 40$ ✓ $\Rightarrow (x + 4)^{2} = 4$ , so $x + 4 = \pm 2$ ✓ $x = - 6$ gives $y = 5 + 3 (- 2) = - 1$ , i.e. $(- 6; - 1)$ in the third quadrant — rejected ✓ $x = - 2$ gives $y = 5 + 3 (2) = 11$ $∴ P (- 2; 11)$ ✓

Modelled on Nov 2024 Paper 2, Question 3 — same skills, new scenario.

Question 4

Circle: centre, tangents & inscribed anglesComplex20 marksPaper 2

In the diagram, $M (3; 2)$ is the centre of a circle. The points $P (- 1; - 1)$ , $R (d; 6)$ and $T (6; - 2)$ , with $d > 0$ , lie on the circle, and $P N$ is a diameter. The tangent to the circle at $P$ is drawn, as well as the chords $T P$ , $T R$ and $T N$ .

aWrite down the coordinates of $N$ . (2)
bDetermine the equation of the tangent to the circle at $P$ in the form $y = m x + c$ . (4)
cShow that the equation of the circle can be written as $x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ . (4)
dShow that $d = 6$ . (2)
eCalculate the size of $P \hat{T} R$ . (5)
fProve that $T P ⊥ T N$ . (3)

This question has a diagram, shown in the app.

Hint

Use the centre as midpoint of the diameter, remember that a tangent is perpendicular to the radius at the point of contact, and find the angle at $T$ from the inclinations of the two chords meeting there.

Worked answer

a) $M$ is the midpoint of the diameter $P N$ , so $x_{N} = 2 (3) - (- 1) = 7$ ✓ and $y_{N} = 2 (2) - (- 1) = 5$ , giving $N (7; 5)$ ✓

b) $m_{M P} = \frac{- 1 - 2}{- 1 - 3} = \frac{3}{4}$ ✓ The tangent is perpendicular to radius $M P$ (tan $⊥$ radius), so $m_{tan} = - \frac{4}{3}$ ✓ $y - (- 1) = - \frac{4}{3} (x - (- 1))$ ✓ $\Rightarrow y = - \frac{4}{3} x - \frac{7}{3}$ ✓

c) $r^{2} = M P^{2} = (- 1 - 3)^{2} + (- 1 - 2)^{2} = 16 + 9 = 25$ ✓ so the circle is $(x - 3)^{2} + (y - 2)^{2} = 25$ ✓ Expanding: $x^{2} - 6 x + 9 + y^{2} - 4 y + 4 = 25$ ✓ $\Rightarrow x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ ✓

d) Substitute $y = 6$ : $x^{2} + 36 - 6 x - 24 - 12 = 0$ ✓ $\Rightarrow x^{2} - 6 x = 0 \Rightarrow x (x - 6) = 0 \Rightarrow x = 0$ or $x = 6$ . Since $d > 0$ : $d = 6$ ✓

e) $m_{T P} = \frac{- 1 - ( - 2 )}{- 1 - 6} = - \frac{1}{7}$ ✓ Inclination of $T P$ : $tan θ = - \frac{1}{7} \Rightarrow θ = 18 0^{\circ} - 8, 1 3^{\circ} = 171, 8 7^{\circ}$ ✓ $T (6; - 2)$ and $R (6; 6)$ have equal $x$ -coordinates, so $T R$ is vertical with inclination $9 0^{\circ}$ ✓ $P \hat{T} R = 171, 8 7^{\circ} - 9 0^{\circ}$ ✓ $= 81, 8 7^{\circ}$ ✓ (Check: the central angle $P \hat{M} R = 163, 7 4^{\circ}$ and the ∠ at centre $= 2 \times$ ∠ at circumf.)

f) $m_{T P} = - \frac{1}{7}$ (from (e)) ✓ $m_{T N} = \frac{5 - ( - 2 )}{7 - 6} = 7$ ✓ $m_{T P} \times m_{T N} = - \frac{1}{7} \times 7 = - 1$ ✓ $∴ T P ⊥ T N$ (equivalently: $P N$ is a diameter, so $P \hat{T} N = 9 0^{\circ}$ , ∠ in semi-circle)

Modelled on Nov 2024 Paper 2, Question 4 — same skills, new scenario.

Question 5

Triangle & parallelogram: surd length, inclination, foot of perpendicular & areaRoutine18 marksPaper 2

In the diagram, $A (- 5; 8)$ , $B (- 4; 1)$ and $C (2; 4)$ are the vertices of $△ A B C$ drawn in a Cartesian plane.

aCalculate the length of $B C$ . Leave your answer in simplest surd form. (2)
bCalculate the gradient of $B C$ . (2)
cDetermine $θ$ , the angle of inclination of $B C$ , correct to TWO decimal places. (2)
dDetermine the equation of the line $B C$ in the form $y = m x + c$ . (2)
eDetermine the coordinates of $D$ if $A B C D$ is a parallelogram. (2)
fThe perpendicular drawn from $A$ to $B C$ meets $B C$ at the point $P$ . Calculate the coordinates of $P$ . (5)
gHence, calculate the area of parallelogram $A B C D$ . (3)

This question has a diagram, shown in the app.

Hint

Every early part works off $B$ and $C$ only; for $D$ shift $A$ by the same amount that takes $B$ to $C$ , and for $P$ solve the equation of $B C$ simultaneously with the perpendicular line through $A$ .

Worked answer

a) $B C = (2 - (- 4))^{2} + (4 - 1)^{2} = 45$ ✓ $= 35 \approx 6, 71$ units ✓

b) $m_{B C} = \frac{4 - 1}{2 - ( - 4 )} = \frac{3}{6}$ ✓ $= \frac{1}{2}$ ✓

c) $tan θ = m_{B C} = \frac{1}{2}$ ✓ $\Rightarrow θ = 26, 5 7^{\circ}$ ✓

d) $y = \frac{1}{2} x + c$ ✓ Substitute $B (- 4; 1)$ : $1 = \frac{1}{2} (- 4) + c$ , so $c = 3$ ✓ Thus $y = \frac{1}{2} x + 3$

e) $A B C D$ is a parallelogram, so $A D = B C$ ✓ giving $D = (- 5 + 2 - (- 4); 8 + 4 - 1) = (1; 11)$ ✓

f) $B C ⊥ A P$ , so $m_{A P} = - 2$ ✓ (product of gradients $= - 1$ ) Line $A P$ : $y - 8 = - 2 (x + 5)$ , i.e. $y = - 2 x - 2$ ✓ At $P$ : $- 2 x - 2 = \frac{1}{2} x + 3$ ✓ $\Rightarrow - \frac{5}{2} x = 5$ , so $x = - 2$ ✓ and $y = \frac{1}{2} (- 2) + 3 = 2$ $∴ P (- 2; 2)$ ✓

g) $A P = (- 2 - (- 5))^{2} + (2 - 8)^{2} = 45 = 35$ ✓ Area of $A B C D = B C \times A P$ ✓ $= 35 \times 35 = 45$ square units ✓

Modelled on Nov 2025 Paper 2, Question 3 — same skills, new scenario.

Question 6

Circle: equation, tangents from an external pointComplex21 marksPaper 2

In the diagram, $M (p; q)$ is the centre of a circle that cuts the $x$ -axis at $A$ and $B$ , and $M$ lies above the $x$ -axis. The vertical line through $M$ crosses the $x$ -axis at $D$ , cuts the circle at $E (3; - 2)$ and runs on to the point $C$ below the circle. $C A$ and $C B$ are tangents to the circle at $A$ and $B$ respectively, and $A D = q - 2$ .

aWrite down the value of $p$ . (1)
bShow that $q = 8$ . (4)
cDetermine the equation of the circle in centre-radius form. (2)
dThe circle is translated 12 units to the right and 3 units down. Calculate the shortest distance between the translated circle and the $y$ -axis. (1)
eCalculate the coordinates of $A$ and $B$ . (3)
fDetermine the equation of the tangent $C A$ in the form $y = m x + c$ . (4)
gCalculate the coordinates of $C$ . (2)
hCalculate the size of $A \hat{C} B$ , the angle between the two tangents. Round off your answer to TWO decimal places. (4)

This question has a diagram, shown in the app.

Hint

Read $p$ from the vertical line through $M$ and $E$ , write the radius $M E$ in terms of $q$ for Pythagoras in $△ M D A$ , and use tangent $⊥$ radius for the gradients, remembering that $C$ lies on the line $x = p$ .

Worked answer

a) $E$ lies on the vertical line through $M$ and $D$ , so $p = 3$ ✓

b) $M E$ is a radius, and $M$ , $D$ and $E$ all lie on the line $x = 3$ , so $r = M E = q - (- 2) = q + 2$ ✓ In $△ M D A$ , $M D = q$ and $M \hat{D} A = 9 0^{\circ}$ , so $M A^{2} = M D^{2} + A D^{2}$ (Pythagoras): $(q + 2)^{2} = q^{2} + (q - 2)^{2}$ ✓ $q^{2} + 4 q + 4 = 2 q^{2} - 4 q + 4 \Rightarrow q^{2} - 8 q = 0 \Rightarrow q (q - 8) = 0$ ✓ $q = 8$ , rejecting $q = 0$ since $M$ lies above the $x$ -axis and $A D = q - 2 > 0$ ✓

c) $r = 8 + 2 = 10$ ✓ so the circle is $(x - 3)^{2} + (y - 8)^{2} = 100$ ✓

d) New centre $(3 + 12; 8 - 3) = (15; 5)$ , so the shortest distance $= 15 - 10 = 5$ units ✓

e) Put $y = 0$ : $(x - 3)^{2} + 64 = 100$ ✓ $(x - 3)^{2} = 36 \Rightarrow x = 3 \pm 6$ ✓ $A (- 3; 0)$ and $B (9; 0)$ ✓

f) $m_{M A} = \frac{8 - 0}{3 - ( - 3 )} = \frac{4}{3}$ ✓ $C A ⊥ M A$ (tan $⊥$ radius), so $m_{C A} = - \frac{3}{4}$ ✓ $y - 0 = - \frac{3}{4} (x + 3)$ ✓ $\Rightarrow y = - \frac{3}{4} x - \frac{9}{4}$ ✓

g) $C$ lies on $C A$ and on the vertical line $x = 3$ ✓ $y = - \frac{3}{4} (3) - \frac{9}{4} = - 4, 5$ , so $C (3; - 4, 5)$ ✓

h) $M \hat{A} C = 9 0^{\circ}$ (tan $⊥$ radius) and $M C = 8 + 4, 5 = 12, 5$ ✓ $sin A \hat{C} M = \frac{M A}{M C} = \frac{10}{12 , 5} = 0, 8$ ✓ $A \hat{C} M = 53, 1 3^{\circ}$ ✓ $C M$ bisects $A \hat{C} B$ ( $C A = C B$ , tangents from a common point), so $A \hat{C} B = 2 \times 53, 1 3^{\circ} = 106, 2 6^{\circ}$ ✓

Modelled on Nov 2025 Paper 2, Question 4 — same skills, new scenario.

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