Secondary 4 Additional Mathematics Geometry Trigonometry Quiz

Secondary 4 Additional Mathematics Quiz - Geometry Trigonometry

Answer Key

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1. Prove: $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$ [3 marks]

Working:

LHS $= \dfrac{\sin 2\theta}{1 + \cos 2\theta}$

Using double-angle identities: $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$

$= \dfrac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)}$

$= \dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}$

$= \dfrac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS} \quad \checkmark$

[Marking notes: 1 mark for correct double-angle substitution, 1 mark for simplification, 1 mark for reaching RHS.]

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2. Solve $2\cos^2 x - 3\cos x + 1 = 0$ for $0° \leq x \leq 360°$ [3 marks]

Working:

Let $u = \cos x$:

$2u^2 - 3u + 1 = 0$

$(2u - 1)(u - 1) = 0$

$u = \frac{1}{2} \quad \text{or} \quad u = 1$

Case 1: $\cos x = \frac{1}{2}$

$x = 60°, \quad x = 300°$

Case 2: $\cos x = 1$

$x = 0°, \quad x = 360°$

Answer: $x = 0°, 60°, 300°, 360°$

[Marking notes: 1 mark for factorisation, 1 mark for correct values from $\cos x = \frac{1}{2}$, 1 mark for including all valid solutions including $0°$ and $360°$. Common mistake: forgetting $0°$ and $360°$ when $\cos x = 1$.]

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3. Express $5\sin\theta - 12\cos\theta$ in the form $R\sin(\theta - \alpha)$ [3 marks]

Working:

$R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$

Comparing with $5\sin\theta - 12\cos\theta$:

$R\cos\alpha = 5, \quad R\sin\alpha = 12$

$R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

$\tan\alpha = \frac{12}{5}$

$\alpha = \tan^{-1}\left(\frac{12}{5}\right) = 67.38° \quad \text{(to 2 d.p.)}$

Answer: $5\sin\theta - 12\cos\theta = 13\sin(\theta - 67.38°)$

[Marking notes: 1 mark for $R = 13$, 1 mark for correct method to find $\alpha$, 1 mark for $\alpha = 67.38°$ and correct form.]

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4. Solve $\tan 2x = \sqrt{3}$ for $0° \leq x \leq 180°$ [3 marks]

Working:

$\tan 2x = \sqrt{3}$

$2x = 60°, 240°, 420°, 600° \quad (\text{since } 0° \leq 2x \leq 360° \text{ gives } 0° \leq x \leq 180°)$

Wait — since $0° \leq x \leq 180°$, then $0° \leq 2x \leq 360°$.

$2x = 60°, 240°$

$x = 30°, 120°$

Answer: $x = 30°, 120°$

[Marking notes: 1 mark for $\tan^{-1}(\sqrt{3}) = 60°$, 1 mark for finding both values of $2x$ in range, 1 mark for correct final answers. Common mistake: not extending to the second solution $2x = 240°$.]

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5. Prove: $\dfrac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta$ [3 marks]

Working:

LHS $= \dfrac{1 - \cos 2\theta}{\sin 2\theta}$

Using identities: $\cos 2\theta = 1 - 2\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$

$= \dfrac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta}$

$= \dfrac{2\sin^2\theta}{2\sin\theta\cos\theta}$

$= \dfrac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS} \quad \checkmark$

[Marking notes: 1 mark for correct identity substitution, 1 mark for simplification, 1 mark for reaching RHS.]

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6. Find equation of $l_2$ through $C(3, 2)$, perpendicular to line through $A(1,5)$ and $B(4,-1)$ [3 marks]

Working:

Gradient of $l_1$:

$m_1 = \frac{-1 - 5}{4 - 1} = \frac{-6}{3} = -2$

Since $l_2 \perp l_1$:

$m_2 = \frac{1}{2} \quad (m_1 \cdot m_2 = -1)$

Equation of $l_2$ through $(3, 2)$:

$y - 2 = \frac{1}{2}(x - 3)$

$2y - 4 = x - 3$

$x - 2y + 1 = 0$

Answer: $x - 2y + 1 = 0$ (or $y = \frac{1}{2}x + \frac{1}{2}$)

[Marking notes: 1 mark for gradient of $l_1$, 1 mark for perpendicular gradient, 1 mark for correct equation.]

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7. Find intersection of $3x + 2y = 12$ and $x - y = 1$ [3 marks]

Working:

From the second equation: $x = y + 1$

Substitute into the first:

$3(y + 1) + 2y = 12$

$3y + 3 + 2y = 12$

$5y = 9$

$y = \frac{9}{5} = 1.8$

$x = 1.8 + 1 = 2.8 = \frac{14}{5}$

Answer: $\left(\dfrac{14}{5}, \dfrac{9}{5}\right)$ or $(2.8, 1.8)$

[Marking notes: 1 mark for substitution, 1 mark for solving, 1 mark for both coordinates.]

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8. Find centre of circle through $A(2,3)$, $B(6,7)$, $C(4,-1)$ [3 marks]

Working:

The centre lies on the perpendicular bisectors of any two chords.

Perpendicular bisector of $AB$:

Midpoint of $AB = \left(\dfrac{2+6}{2}, \dfrac{3+7}{2}\right) = (4, 5)$

Gradient of $AB = \dfrac{7-3}{6-2} = 1$, so perpendicular gradient $= -1$

Equation: $y - 5 = -(x - 4)$, i.e. $y = -x + 9$ ... (i)

Perpendicular bisector of $BC$:

Midpoint of $BC = \left(\dfrac{6+4}{2}, \dfrac{7+(-1)}{2}\right) = (5, 3)$

Gradient of $BC = \dfrac{-1-7}{4-6} = \dfrac{-8}{-2} = 4$, so perpendicular gradient $= -\dfrac{1}{4}$

Equation: $y - 3 = -\dfrac{1}{4}(x - 5)$, i.e. $4y - 12 = -x + 5$, so $x + 4y = 17$ ... (ii)

Solving (i) and (ii):

Substitute $x = 9 - y$ into (ii):

$(9 - y) + 4y = 17$

$3y = 8$

$y = \dfrac{8}{3}$

$x = 9 - \dfrac{8}{3} = \dfrac{19}{3}$

Answer: Centre $= \left(\dfrac{19}{3}, \dfrac{8}{3}\right)$

[Marking notes: 1 mark for finding one perpendicular bisector correctly, 1 mark for finding the second, 1 mark for solving simultaneously.]

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9. Circle with centre $(3, -2)$ through $(7, 1)$ [4 marks]

(a) Radius:

$r = \sqrt{(7-3)^2 + (1-(-2))^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Answer: $r = 5$

(b) Equation:

$(x - 3)^2 + (y + 2)^2 = 25$

(c) Distance from $(0, 2)$ to centre $(3, -2)$:

$d = \sqrt{(0-3)^2 + (2-(-2))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Since $d = r = 5$, the point $(0, 2)$ lies on the circle.

[Marking notes: 1 mark for (a), 1 mark for (b), 1 mark for distance calculation in (c), 1 mark for correct conclusion with justification.]

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10. Find $k$ such that $y = 2x + k$ is tangent to $x^2 + y^2 = 25$ [4 marks]

Working:

Substitute $y = 2x + k$ into the circle:

$x^2 + (2x + k)^2 = 25$

$x^2 + 4x^2 + 4kx + k^2 = 25$

$5x^2 + 4kx + (k^2 - 25) = 0$

For tangency, the discriminant $= 0$:

$(4k)^2 - 4(5)(k^2 - 25) = 0$

$16k^2 - 20k^2 + 500 = 0$

$-4k^2 + 500 = 0$

$k^2 = 125$

$k = \pm 5\sqrt{5}$

Answer: $k = 5\sqrt{5}$ or $k = -5\sqrt{5}$

[Marking notes: 1 mark for correct substitution, 1 mark for forming the quadratic, 1 mark for setting discriminant to zero, 1 mark for correct values of $k$.]

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11. Circle with diameter endpoints $P(-1, 2)$ and $Q(5, 8)$ [4 marks]

(a) Centre $=$ midpoint of $PQ$:

$\text{Centre} = \left(\dfrac{-1+5}{2}, \dfrac{2+8}{2}\right) = (2, 5)$

(b) Radius:

$r = \frac{1}{2}\sqrt{(5-(-1))^2 + (8-2)^2} = \frac{1}{2}\sqrt{36 + 36} = \frac{1}{2}\sqrt{72} = 3\sqrt{2}$

Equation:

$(x - 2)^2 + (y - 5)^2 = 18$

(c) Gradient of radius to $P$:

$m_{CP} = \frac{2 - 5}{-1 - 2} = \frac{-3}{-3} = 1$

Gradient of tangent $= -1$ (perpendicular)

Equation through $P(-1, 2)$:

$y - 2 = -(x + 1)$

$y = -x + 1$

Answer: $x + y - 1 = 0$

[Marking notes: 1 mark for (a), 1 mark for (b) equation, 1 mark for gradient of radius in (c), 1 mark for tangent equation in (c).]

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12. Perpendicular distance and foot of perpendicular from $A(1, -3)$ to $4x - 3y + 6 = 0$ [4 marks]

(a) Perpendicular distance:

$d = \frac{|4(1) - 3(-3) + 6|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 + 9 + 6|}{\sqrt{25}} = \frac{19}{5} = 3.8$

(b) The foot of the perpendicular lies on the line through $A$ perpendicular to $l$.

Gradient of $l$: $y = \frac{4}{3}x + 2$, so $m = \frac{4}{3}$

Perpendicular gradient $= -\frac{3}{4}$

Equation through $A(1, -3)$:

$y + 3 = -\frac{3}{4}(x - 1)$

$4y + 12 = -3x + 3$

$3x + 4y + 9 = 0 \quad \text{...(i)}$

Solve simultaneously with $4x - 3y + 6 = 0$ ... (ii):

From (i) $\times 3$: $9x + 12y + 27 = 0$

From (ii) $\times 4$: $16x - 12y + 24 = 0$

Adding: $25x + 51 = 0$, so $x = -\dfrac{51}{25}$

From (ii): $4\left(-\dfrac{51}{25}\right) - 3y + 6 = 0$

$-\dfrac{204}{25} + 6 = 3y$

$\dfrac{-204 + 150}{25} = 3y$

$-\dfrac{54}{25} = 3y$

$y = -\dfrac{18}{25}$

Answer: Foot $= \left(-\dfrac{51}{25}, -\dfrac{18}{25}\right)$

[Marking notes: 1 mark for (a) correct formula and answer, 1 mark for perpendicular line equation in (b), 1 mark for solving, 1 mark for correct coordinates.]

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13. Triangle $PQR$: $PQ = 8$, $QR = 11$, $\angle PQR = 52°$ [4 marks]

(a) Using the cosine rule on side $PR$ (opposite $\angle Q$):

$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)$

$PR^2 = 64 + 121 - 2(8)(11)\cos 52°$

$PR^2 = 185 - 176 \times 0.6157$

$PR^2 = 185 - 108.36 = 76.64$

$PR = \sqrt{76.64} = 8.75 \text{ cm (3 s.f.)}$

(b) Area:

$\text{Area} = \frac{1}{2}(PQ)(QR)\sin(\angle PQR)$

$= \frac{1}{2}(8)(11)\sin 52°$

$= 44 \times 0.7880 = 34.7 \text{ cm}^2 \text{ (3 s.f.)}$

[Marking notes: 1 mark for correct cosine rule setup in (a), 1 mark for correct answer in (a), 1 mark for correct area formula in (b), 1 mark for correct answer in (b).]

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14. Height of building from angles of elevation [4 marks]

Working:

Let the height of the building be $h$ m, and let the distance from $A$ to the base be $d$ m.

From point $A$: $\tan 35° = \dfrac{h}{d}$, so $h = d\tan 35°$ ... (i)

From point $B$: $\tan 20° = \dfrac{h}{d + 40}$, so $h = (d + 40)\tan 20°$ ... (ii)

Equating:

$d\tan 35° = (d + 40)\tan 20°$

$d(0.7002) = d(0.3640) + 14.56$

$0.3362d = 14.56$

$d = 43.31 \text{ m}$

$h = 43.31 \times \tan 35° = 43.31 \times 0.7002 = 30.3 \text{ m (3 s.f.)}$

Answer: Height $= 30.3$ m

[Marking notes: 1 mark for setting up two equations, 1 mark for equating, 1 mark for solving for $d$, 1 mark for finding $h$.]

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15. Triangle $ABC$: $AB = 12$, $AC = 9$, $\angle BAC = 68°$, $AD$ bisects $\angle BAC$ [4 marks]

(a) Using the cosine rule:

$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)$

$BC^2 = 144 + 81 - 2(12)(9)\cos 68°$

$BC^2 = 225 - 216 \times 0.3746$

$BC^2 = 225 - 80.91 = 144.09$

$BC = 12.0 \text{ cm (3 s.f.)}$

(b) By the angle bisector theorem:

$\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{9} = \frac{4}{3}$

Answer: $BD : DC = 4 : 3$

[Marking notes: 1 mark for correct cosine rule setup in (a), 1 mark for correct answer in (a), 1 mark for angle bisector theorem in (b), 1 mark for correct ratio in (b).]

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16. Height of tower from two angles of elevation [4 marks]

(a) Let the distance from $P$ to the base of the tower be $d$ m.

$h = d\tan 48°$

Also from $Q$: the distance from $Q$ to the base is $(d - 30)$ m (since $Q$ is further from the tower if the angle is smaller — actually, since the angle at $Q$ is smaller, $Q$ is further away, so distance from $Q$ to base $= d + 30$).

$h = (d + 30)\tan 32°$

(b) Equating:

$d\tan 48° = (d + 30)\tan 32°$

$d(1.1106) = d(0.6249) + 18.747$

$0.4857d = 18.747$

$d = 38.60 \text{ m}$

$h = 38.60 \times \tan 48° = 38.60 \times 1.1106 = 42.9 \text{ m (3 s.f.)}$

Answer: Height $= 42.9$ m

[Marking notes: 1 mark for correct expression in (a), 1 mark for setting up equation in (b), 1 mark for solving, 1 mark for correct height.]

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17. Quadrilateral $ABCD$: $AB = 6$, $BC = 8$, $CD = 5$, $DA = 7$, $\angle ABC = 110°$, $\angle ACD = 40°$ [5 marks]

(a) Using the cosine rule in $\triangle ABC$:

$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$

$AC^2 = 36 + 64 - 2(6)(8)\cos 110°$

$AC^2 = 100 - 96 \times (-0.3420)$

$AC^2 = 100 + 32.83 = 132.83$

$AC = 11.5 \text{ cm (3 s.f.)}$

(b) Area of $\triangle ABC$:

$\text{Area} = \frac{1}{2}(AB)(BC)\sin(\angle ABC) = \frac{1}{2}(6)(8)\sin 110°$

$= 24 \times 0.9397 = 22.6 \text{ cm}^2 \text{ (3 s.f.)}$

(c) In $\triangle ACD$, we need another angle. Using the sine rule in $\triangle ACD$:

We know $AC = 11.53$, $CD = 5$, $DA = 7$, and $\angle ACD = 40°$.

Using the sine rule to find $\angle CAD$:

$\frac{CD}{\sin(\angle CAD)} = \frac{DA}{\sin(\angle ACD)}$

$\frac{5}{\sin(\angle CAD)} = \frac{7}{\sin 40°}$

$\sin(\angle CAD) = \frac{5 \times \sin 40°}{7} = \frac{5 \times 0.6428}{7} = 0.4591$

$\angle CAD = 27.34°$

$\angle ADC = 180° - 40° - 27.34° = 112.66°$

Area of $\triangle ACD$:

$= \frac{1}{2}(AC)(CD)\sin(\angle ACD) = \frac{1}{2}(11.53)(5)\sin 40°$

$= 28.825 \times 0.6428 = 18.5 \text{ cm}^2 \text{ (3 s.f.)}$

(d) Total area:

$= 22.6 + 18.5 = 41.1 \text{ cm}^2 \text{ (3 s.f.)}$

[Marking notes: 1 mark for (a), 1 mark for (b), 1 mark for finding an angle in (c), 1 mark for area of triangle ACD in (c), 1 mark for total in (d).]

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18. Two ships leaving port [5 marks]

(a) After 2 hours:

Ship $X$: distance $= 15 \times 2 = 30$ km

Ship $Y$: distance $= 20 \times 2 = 40$ km

(b) The angle between their paths $= 145° - 55° = 90°$

Using the cosine rule (or Pythagoras since angle $= 90°$):

$XY^2 = 30^2 + 40^2 = 900 + 1600 = 2500$

$XY = 50.0 \text{ km}$

(c) To find the bearing of $X$ from $Y$:

Consider the triangle. Place $P$ at origin. Ship $X$ is at bearing $055°$ from $P$, distance 30 km. Ship $Y$ is at bearing $145°$ from $P$, distance 40 km.

The angle $\angle XPY = 90°$.

In triangle $XPY$, $\tan(\angle PYX) = \dfrac{PX}{PY} = \dfrac{30}{40} = 0.75$

$\angle PYX = 36.87°$

The bearing of $X$ from $Y$: From $Y$, the direction to $P$ is $145° + 180° = 325°$. Then turn by angle $\angle PYX = 36.87°$ towards $X$.

Bearing of $X$ from $Y = 325° - 36.87° = 288°$ (to nearest degree)

Alternative method using coordinates:

$X$: $(30\sin 55°, 30\cos 55°) = (24.576, 17.207)$

$Y$: $(40\sin 145°, 40\cos 145°) = (22.943, -32.766)$

Vector from $Y$ to $X$: $(24.576 - 22.943, 17.207 - (-32.766)) = (1.633, 49.973)$

Bearing $= \tan^{-1}\left(\dfrac{1.633}{49.973}\right) = \tan^{-1}(0.03268) = 1.87°$

Since $x > 0$ and $y > 0$, bearing $= 002°$ (to nearest degree)

Answer: Bearing of $X$ from $Y = 002°$

[Marking notes: 1 mark for (a), 1 mark for (b), 1 mark for correct method in (c), 1 mark for correct angle calculation in (c), 1 mark for correct bearing in (c).]

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19. Graph of $y = a\sin(bx) + c$ through $(0, 1)$, $\left(\frac{\pi}{4}, 3\right)$, $\left(\frac{\pi}{2}, 1\right)$ [5 marks]

(a) From $(0, 1)$: $a\sin(0) + c = 1$, so $c = 1$.

From $\left(\frac{\pi}{2}, 1\right)$: $a\sin\left(\frac{b\pi}{2}\right) + 1 = 1$, so $\sin\left(\frac{b\pi}{2}\right) = 0$.

This gives $\frac{b\pi}{2} = \pi$ (smallest positive), so $b = 2$.

From $\left(\frac{\pi}{4}, 3\right)$: $a\sin\left(\frac{2\pi}{4}\right) + 1 = 3$

$a\sin\left(\frac{\pi}{2}\right) + 1 = 3$

$a(1) = 2$, so $a = 2$.

Answer: $a = 2$, $b = 2$, $c = 1$

(b) Amplitude $= |a| = 2$

Period $= \dfrac{2\pi}{b} = \dfrac{2\pi}{2} = \pi$

Maximum value $= c + a = 1 + 2 = 3$

(c) The graph of $y = 2\sin(2x) + 1$ for $0 \leq x \leq 2\pi$:

• Starts at $(0, 1)$, rises to maximum $\left(\dfrac{\pi}{4}, 3\right)$, returns to $\left(\dfrac{\pi}{2}, 1\right)$, drops to minimum $\left(\dfrac{3\pi}{4}, -1\right)$, returns to $(\pi, 1)$, rises to $\left(\dfrac{5\pi}{4}, 3\right)$, returns to $\left(\dfrac{3\pi}{2}, 1\right)$, drops to $\left(\dfrac{7\pi}{4}, -1\right)$, returns to $(2\pi, 1)$.

[Marking notes: 1 mark for $c = 1$, 1 mark for $b = 2$, 1 mark for $a = 2$, 1 mark for amplitude and period in (b), 1 mark for correct sketch in (c) with key points labelled.]

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20. Circle $x^2 + y^2 - 6x + 4y - 12 = 0$ and second circle centre $(10, -3)$, radius 5 [5 marks]

(a) Complete the square:

$x^2 - 6x + y^2 + 4y = 12$

$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$

$(x - 3)^2 + (y + 2)^2 = 25$

Centre $= (3, -2)$, radius $= 5$

(b) The chord has midpoint $(4, -1)$. The line from the centre to the midpoint is perpendicular to the chord.

Gradient of line from centre $(3, -2)$ to midpoint $(4, -1)$:

$m = \frac{-1 - (-2)}{4 - 3} = \frac{1}{1} = 1$

Gradient of chord $= -1$ (perpendicular)

Equation through $(4, -1)$:

$y + 1 = -(x - 4)$

$y = -x + 3$

Answer: $x + y - 3 = 0$

(c) Distance between centres:

$d = \sqrt{(10 - 3)^2 + (-3 - (-2))^2} = \sqrt{49 + 1} = \sqrt{25} = 5$

Wait: $\sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \approx 7.07$

Sum of radii $= 5 + 5 = 10$

Difference of radii $= 5 - 5 = 0$

Since $d = \sqrt{50} \approx 7.07$ and this is between $0$ and $10$, the circles intersect at two points (they do not touch).

Let me recalculate: $(10-3)^2 + (-3+2)^2 = 49 + 1 = 50$, so $d = \sqrt{50} = 5\sqrt{2} \approx 7.07$.

Since $|r_1 - r_2| = 0 < d = 7.07 < r_1 + r_2 = 10$, the circles intersect at two distinct points.

Correction: The circles do not touch. They intersect at two points.

Answer: The circles intersect at two points (they do not touch), since the distance between centres $\sqrt{50} \approx 7.07$ lies strictly between $|r_1 - r_2| = 0$ and $r_1 + r_2 = 10$.

[Marking notes: 1 mark for completing the square in (a), 1 mark for centre and radius in (a), 1 mark for correct gradient and equation in (b), 1 mark for distance calculation in (c), 1 mark for correct conclusion with justification in (c).]

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