Secondary 4 Additional Mathematics Practice Paper 3

TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

Answer Key and Marking Scheme

Version: 3 of 5
Total Marks: 100

Section A: Coordinate Geometry Fundamentals (25 marks)

1. Find the coordinates of the point that divides $A(-3, 7)$ and $B(5, -1)$ internally in the ratio $3:1$. [3 marks]

Method: Use the section formula: if $P$ divides $AB$ in ratio $m:n$, then $P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$

Working: $P = \left(\frac{3(5) + 1(-3)}{3+1}, \frac{3(-1) + 1(7)}{4}\right)$

$= \left(\frac{15-3}{4}, \frac{-3+7}{4}\right)$

$= \left(\frac{12}{4}, \frac{4}{4}\right)$

Answer: $(3, 1)$ [3 marks]

Marking: Formula or method (1), correct substitution (1), final answer (1).

2. The points $P(2, -1)$, $Q(4, 3)$ and $R(6, k)$ are collinear. [3 marks]

Method: Collinear points have equal gradients. Gradient of $PQ$ = gradient of $QR$.

Working: $\text{Gradient of } PQ = \frac{3-(-1)}{4-2} = \frac{4}{2} = 2$

$\text{Gradient of } QR = \frac{k-3}{6-4} = \frac{k-3}{2}$

For collinearity: $\frac{k-3}{2} = 2$

$k - 3 = 4$

Answer: $k = 7$ [3 marks]

Marking: Gradient formula or method (1), correct gradients (1), solving for $k$ (1).

Common error: Using $PR$ instead of consecutive points; still valid but more complex.

3. Perpendicular bisector of $A(-2, 5)$ and $B(4, -3)$. [4 marks]

Method: Need midpoint and perpendicular gradient.

Working: $\text{Midpoint} = \left(\frac{-2+4}{2}, \frac{5+(-3)}{2}\right) = (1, 1)$

$\text{Gradient of } AB = \frac{-3-5}{4-(-2)} = \frac{-8}{6} = -\frac{4}{3}$

$\text{Perpendicular gradient} = \frac{3}{4}$ (negative reciprocal)

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

$4(y-1) = 3(x-1)$

$4y - 4 = 3x - 3$

Answer: $3x - 4y + 1 = 0$ [4 marks]

Marking: Midpoint (1), gradient of AB (1), perpendicular gradient (1), equation manipulation (1).

4. Line $2x - 3y + 5 = 0$

(a) Coordinates of $P$ (on $x$-axis, $y=0$) and $Q$ (on $y$-axis, $x=0$) [2 marks]

Working: For $P$: $2x + 5 = 0 \Rightarrow x = -\frac{5}{2}$, so $P\left(-\frac{5}{2}, 0\right)$

For $Q$: $-3y + 5 = 0 \Rightarrow y = \frac{5}{3}$, so $Q\left(0, \frac{5}{3}\right)$

Answer: $P\left(-2.5, 0\right)$, $Q\left(0, \frac{5}{3}\right)$ [2 marks]

(b) Area of triangle $OPQ$ [2 marks]

Working: $\text{Area} = \frac{1}{2} \times \left|-\frac{5}{2}\right| \times \frac{5}{3} = \frac{1}{2} \times \frac{5}{2} \times \frac{5}{3} = \frac{25}{12}$

Answer: $\frac{25}{12}$ or $2.08$ units² [2 marks]

(c) Perpendicular distance from origin to line [2 marks]

Method: Use formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}$ where line is $ax+by+c=0$.

Working: $d = \frac{|2(0) - 3(0) + 5|}{\sqrt{4+9}} = \frac{5}{\sqrt{13}} = \frac{5\sqrt{13}}{13}$

Answer: $\frac{5\sqrt{13}}{13}$ or $1.39$ units [2 marks]

5. Line through $(-1, 4)$ parallel to $3x + 2y - 7 = 0$. [3 marks]

Method: Parallel lines have same gradient.

Working: $3x + 2y - 7 = 0 \Rightarrow y = -\frac{3}{2}x + \frac{7}{2}$

Gradient $= -\frac{3}{2}$

Equation: $y - 4 = -\frac{3}{2}(x - (-1)) = -\frac{3}{2}(x+1)$

$2(y-4) = -3(x+1)$

$2y - 8 = -3x - 3$

Answer: $3x + 2y - 5 = 0$ [3 marks]

6. Circle $x^2 + y^2 - 6x + 4y - 12 = 0$

(a) Centre and radius [3 marks]

Method: Complete the square.

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

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

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

Answer: Centre $C(3, -2)$, radius $r = 5$ [3 marks]

Marking: Completing square for $x$ (1), for $y$ (1), centre and radius (1).

(b) Position of point $(5, 2)$ [2 marks]

Working: Distance from centre: $\sqrt{(5-3)^2 + (2-(-2))^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47$

Since $2\sqrt{5} < 5$ (or $20 < 25$):

Answer: The point lies inside the circle [2 marks]

7. Line with gradient $-\frac{2}{5}$ through intersection of $x + 2y = 7$ and $3x - y = 6$. [4 marks]

Working: From second equation: $y = 3x - 6$

Substitute: $x + 2(3x-6) = 7$

$x + 6x - 12 = 7$

$7x = 19$, so $x = \frac{19}{7}$

$y = 3(\frac{19}{7}) - 6 = \frac{57-42}{7} = \frac{15}{7}$

Point of intersection: $\left(\frac{19}{7}, \frac{15}{7}\right)$

Equation: $y - \frac{15}{7} = -\frac{2}{5}\left(x - \frac{19}{7}\right)$

Multiply by 35: $35y - 75 = -14(x - \frac{19}{7}) = -14x + 38$

$14x + 35y - 113 = 0$ or simplify: check arithmetic...

Actually: $14x + 35y = 14(\frac{19}{7}) + 35(\frac{15}{7}) = 38 + 75 = 113$ ✓

Answer: $14x + 35y - 113 = 0$ or $2x + 5y - \frac{113}{7} = 0$...

Better: multiply original by 35: $35y - 75 = -14x + 38$ $14x + 35y - 113 = 0$

Can divide by... gcd(14,35,113)=1, so this is simplest.

Or: $y = -\frac{2}{5}x + c$ with point $\left(\frac{19}{7}, \frac{15}{7}\right)$:

$\frac{15}{7} = -\frac{2}{5} \cdot \frac{19}{7} + c = -\frac{38}{35} + c$

$c = \frac{75+38}{35} = \frac{113}{35}$

So $y = -\frac{2}{5}x + \frac{113}{35}$

Or $35y = -14x + 113$

Answer: $14x + 35y = 113$ or equivalent [4 marks]

Section A Total: 25 marks

Section B: Curves, Tangents and Applications (40 marks)

8. Curve $y = x^2 - 5x + 6$ intersects line $y = x - 1$ at $A$ and $B$. [4 marks]

Working: $x^2 - 5x + 6 = x - 1$

$x^2 - 6x + 7 = 0$

$x = \frac{6 \pm \sqrt{36-28}}{2} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}$

When $x = 3 + \sqrt{2}$: $y = 3 + \sqrt{2} - 1 = 2 + \sqrt{2}$

When $x = 3 - \sqrt{2}$: $y = 3 - \sqrt{2} - 1 = 2 - \sqrt{2}$

Answer: $A(3+\sqrt{2}, 2+\sqrt{2})$, $B(3-\sqrt{2}, 2-\sqrt{2})$ [4 marks]

Or either order. Approx: $A(4.41, 3.41)$, $B(1.59, 0.586)$

Marking: Setting up equation (1), solving quadratic (1), both $x$-values (1), both $y$-values (1).

9. Parabola $y = x^2 - 4x + 3$

(a) Coordinates of $A$, $B$, $C$ [3 marks]

Working: $x$-intercepts: $x^2 - 4x + 3 = 0$

$(x-1)(x-3) = 0$, so $x = 1$ or $x = 3$

$A(1, 0)$ and $B(3, 0)$

Vertex: $x = -\frac{b}{2a} = \frac{4}{2} = 2$

$y = 4 - 8 + 3 = -1$

Answer: $A(1, 0)$, $B(3, 0)$, $C(2, -1)$ [3 marks]

(b) Tangent at $D(4, 3)$ [3 marks]

Working: $\frac{dy}{dx} = 2x - 4$

At $x = 4$: gradient $= 8 - 4 = 4$

Equation: $y - 3 = 4(x - 4)$

$y = 4x - 16 + 3 = 4x - 13$

Answer: $y = 4x - 13$ or $4x - y - 13 = 0$ [3 marks]

(c) Point $E$ and area of triangle $CDE$ [4 marks]

Working: Tangent meets $x$-axis: $0 = 4x - 13$, so $x = \frac{13}{4}$

$E\left(\frac{13}{4}, 0\right)$

Area of triangle $CDE$ with $C(2, -1)$, $D(4, 3)$, $E(\frac{13}{4}, 0)$

Using formula: $\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

$= \frac{1}{2}|2(3-0) + 4(0-(-1)) + \frac{13}{4}(-1-3)|$

$= \frac{1}{2}|6 + 4 + \frac{13}{4}(-4)|$

$= \frac{1}{2}|10 - 13|$

$= \frac{1}{2}|-3| = \frac{3}{2}$

Answer: $E\left(\frac{13}{4}, 0\right)$ or $(3.25, 0)$, Area $= \frac{3}{2}$ or $1.5$ units² [4 marks]

Marking: Finding E (2), area formula/method (1), final area (1).

10. Curve $y = 2x^3 - 9x^2 + 12x + 5$

(a) Find $\frac{dy}{dx}$ [2 marks]

Answer: $\frac{dy}{dx} = 6x^2 - 18x + 12$ or $6(x^2 - 3x + 2)$ [2 marks]

(b) Stationary points [3 marks]

Working: $6x^2 - 18x + 12 = 0$

$x^2 - 3x + 2 = 0$

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

$x = 1$: $y = 2 - 9 + 12 + 5 = 10$, so $(1, 10)$

$x = 2$: $y = 16 - 36 + 24 + 5 = 9$, so $(2, 9)$

Answer: $(1, 10)$ and $(2, 9)$ [3 marks]

(c) Nature of stationary points [3 marks]

Working: $\frac{d^2y}{dx^2} = 12x - 18$

At $x = 1$: $12 - 18 = -6 < 0$, so maximum at $(1, 10)$

At $x = 2$: $24 - 18 = 6 > 0$, so minimum at $(2, 9)$

Answer: $(1, 10)$ is a maximum, $(2, 9)$ is a minimum [3 marks]

Marking: Second derivative (1), each conclusion with reason (1 each).

Alternative: First derivative test acceptable with clear reasoning.

11. Normal to $y = \frac{1}{x}$ at $x = 2$

(a) Equation of normal [4 marks]

Working: Point: $(2, \frac{1}{2})$

$\frac{dy}{dx} = -\frac{1}{x^2}$, so at $x=2$: gradient of tangent $= -\frac{1}{4}$

Gradient of normal $= 4$ (negative reciprocal)

Equation: $y - \frac{1}{2} = 4(x - 2)$

$y = 4x - 8 + \frac{1}{2} = 4x - \frac{15}{2}$

Or: $2y = 8x - 15$, so $8x - 2y - 15 = 0$

Answer: $y = 4x - \frac{15}{2}$ or $8x - 2y - 15 = 0$ [4 marks]

(b) Normal meets curve again [4 marks]

Working: Substitute into $y = \frac{1}{x}$:

$\frac{1}{x} = 4x - \frac{15}{2}$

Multiply by $2x$: $2 = 8x^2 - 15x$

$8x^2 - 15x - 2 = 0$

$(8x + 1)(x - 2) = 0$

We know $x = 2$ is the original point.

So $x = -\frac{1}{8}$

$y = \frac{1}{-\frac{1}{8}} = -8$

Answer: $\left(-\frac{1}{8}, -8\right)$ [4 marks]

Marking: Substitution (1), rearranging to quadratic (1), solving (1), coordinates (1).

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

(a) Centre [1 mark]

Answer: $C(2, -3)$ [1 mark]

(b) Tangent length from $P(10, 1)$ [3 marks]

Working: Distance $CP = \sqrt{(10-2)^2 + (1-(-3))^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$

Using tangent-secant theorem: $PT^2 = CP^2 - r^2 = 80 - 25 = 55$

Answer: Tangent length $= \sqrt{55}$ or $7.42$ units [3 marks]

(c) Verify $3x - 4y - 16 = 0$ is tangent, find point of contact [4 marks]

Working: Normal form: distance from centre to line equals radius.

Distance from $(2, -3)$ to $3x - 4y - 16 = 0$:

$d = \frac{|3(2) - 4(-3) - 16|}{\sqrt{9+16}} = \frac{|6 + 12 - 16|}{5} = \frac{2}{5}$

Wait... this gives $\frac{2}{5} \neq 5$. Let me recheck.

Actually: $|6 + 12 - 16| = |2| = 2$, and $\sqrt{25} = 5$, so $d = \frac{2}{5}$.

This is not equal to 5. Let me recheck the line equation.

Hmm, problem statement says "Verify that the line... is tangent". Perhaps I need to adjust. Let me check if I misread - maybe the line should be different, or perhaps I made an error.

Wait - let me re-examine. Perhaps the intended line was different. Given the instruction to have valid answers, let me assume the line is actually tangent and I should verify: for a line to be tangent from outside, we'd need the distance to equal 5.

Let's try $3x - 4y + c = 0$ where distance is 5:

$\frac{|6 + 12 + c|}{5} = 5$, so $|18+c| = 25$

$c = 7$ or $c = -43$

So $3x - 4y + 7 = 0$ or $3x - 4y - 43 = 0$ would be tangent.

Given the problem as stated with $3x - 4y - 16 = 0$, let me verify: this is NOT tangent. I should flag this or adjust.

Actually, re-reading - the problem as given in the exam says to verify it. Perhaps there's an error in my transcription or the problem intends for students to discover it's not tangent? Let me re-interpret: "Verify that... is tangent" implies it IS tangent. Let me recheck my distance.

$d = \frac{|6+12-16|}{5} = \frac{2}{5} \neq 5$.

Actually, let me re-examine: $3(2) = 6$, $-4(-3) = +12$, so $6 + 12 - 16 = 2$. Yes, $|2|/5 = 0.4$.

This line is not tangent. I'll adjust my answer to show this, or perhaps the intended line was $3x-4y-43=0$ or $3x-4y+7=0$.

Given I must provide a valid answer key, I'll note: If the line were tangent, the distance would equal 5. For the given line, $d = 0.4 \neq 5$, so this line is not tangent to the circle.

However, to make this a valid problem: if the equation were $3x - 4y + 7 = 0$:

$d = \frac{|6+12+7|}{5} = \frac{25}{5} = 5$ ✓

Point of contact: along normal from centre. Direction of normal is $(3, -4)$, unit vector is $(3/5, -4/5)$.

Point of contact = $(2, -3) + 5 \times (3/5, -4/5) = (2+3, -3-4) = (5, -7)$

Or in opposite direction for other tangent: $(2-3, -3+4) = (-1, 1)$

For $3x-4y+7=0$: the normal direction from centre to line is $(3/5, -4/5)$ or check sign. Actually centre should be in side where $3(2)-4(-3)+7 = 6+12+7=25>0$. The gradient points this way.

Let me provide answer for corrected version or explain.

Given I need to be consistent with the exam paper, I'll provide the methodology but note: If the line is indeed tangent, distance from centre equals radius. For $3x-4y-16=0$, this fails.

Assuming the problem meant $3x - 4y - 43 = 0$: $d = \frac{|6+12-43|}{5} = \frac{25}{5} = 5$ ✓

Point of contact: $(2, -3) + 5 \times (\frac{3}{5}, \frac{-4}{5}) \times$ sign factor.

Since $6+12-43 = -25 < 0$, opposite to $(3,-4)$ direction, so add $5 \times (-\frac{3}{5}, \frac{4}{5}) = (-3, 4)$?

Actually, from centre move along normal by distance 5 towards line. The direction is $-(3,-4)/5 \times 5 = (-3,4)$? Check: new point $(2-3, -3+4) = (-1, 1)$. Verify: $3(-1)-4(1)-43 = -3-4-43 = -50 \neq 0$.

Move the other way: $(2+3, -3-4) = (5, -7)$. Verify: $15+28-43 = 0$ ✓

So point of contact is $(5, -7)$.

Given this complexity, I'll provide:

Revised Answer approach: For the given line $3x-4y-16=0$, distance is $\frac{2}{5} \neq 5$, so it is not tangent. [Students should show this calculation]

If we consider the tangent line with same normal direction, it would be $3x - 4y - 43 = 0$ with point of contact $(5, -7)$.

Marking adjustment: Award marks for correct distance formula application and conclusion.

Given this issue, let me continue with a cleaner track. I'll provide the answer as:

For the given line: Calculate distance and conclude not tangent, or if typo assumed, show method.

Working (assuming intended tangent): Point of contact is where radius to point is perpendicular to tangent, i.e., along normal direction $(3, -4)$.

From centre $(2, -3)$, parameterise: $(2+3t, -3-4t)$

Substitute into circle: $(3t)^2 + (-4t)^2 = 25$, so $9t^2 + 16t^2 = 25t^2 = 25$, thus $t = \pm 1$

Points: $(5, -7)$ and $(-1, 1)$

Check in line $3x-4y-16=0$: for $(5,-7)$: $15+28-16=27\neq 0$. For $(-1,1)$: $-3-4-16=-23\neq 0$.

For $3x-4y-43=0$: $(5,-7)$ gives $15+28-43=0$ ✓

Answer: Assuming corrected equation $3x-4y-43=0$ or $3x-4y+7=0$: point of contact is $(5, -7)$ or $(-1, 1)$ respectively [4 marks]

I'll flag this for version 4 correction. Continuing with paper...

13. Rectangle $A(1,1)$, $B(5,1)$, $C(5,4)$, $D(1,4)$

(a) Diagonal $AC$ [2 marks]

Working: Gradient of $AC = \frac{4-1}{5-1} = \frac{3}{4}$

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

$4y - 4 = 3x - 3$

Answer: $3x - 4y + 1 = 0$ or $y = \frac{3}{4}x + \frac{1}{4}$ [2 marks]

(b) Line through midpoint of $AC$, perpendicular to $AC$ [3 marks]

Working: Midpoint of $AC = (3, 2.5)$

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

Equation: $y - 2.5 = -\frac{4}{3}(x - 3)$

$3y - 7.5 = -4(x-3) = -4x + 12$

$3y = -4x + 19.5$, or $6y = -8x + 39$, or $y = -\frac{4}{3}x + \frac{13}{2}$... let me recheck.

$3y = -4x + 12 + 7.5 = -4x + 19.5 = -\frac{39}{2}$, so $y = -\frac{4}{3}x + \frac{13}{2}$?

Check: $y = -\frac{4}{3}(3) + \frac{13}{2} = -4 + 6.5 = 2.5$ ✓

So $m = -\frac{4}{3}$, $c = \frac{13}{2}$ or $6.5$

Answer: $m = -\frac{4}{3}$, $c = \frac{13}{2}$ or $6.5$ [3 marks]

(c) Verify through midpoint of $BD$ [3 marks]

Working: Midpoint of $BD = \left(\frac{5+1}{2}, \frac{4+1}{2}\right) = (3, 2.5)$

This equals midpoint of $AC$ (as found above).

Explanation: In any rectangle, the diagonals bisect each other. This is a defining property of parallelograms (and rectangles are special parallelograms). The line through the common midpoint perpendicular to one diagonal is the perpendicular bisector of that diagonal, and by symmetry of the rectangle, this line is also the perpendicular bisector of the other diagonal.

Or more directly: since both diagonals share the same midpoint, any line through this midpoint is a line through the midpoint of both diagonals.

Answer: Midpoint of $BD$ is $(3, 2.5)$ which equals midpoint of $AC$. Any line through this point passes through both midpoints. This is true for all rectangles because diagonals of a parallelogram bisect each other. [3 marks]

14. Parabola through $(1,0)$, $(2,3)$, $(-1,6)$

(a) System of equations [2 marks]

Working: $(1,0)$: $a + b + c = 0$

$(2,3)$: $4a + 2b + c = 3$

$(-1,6)$: $a - b + c = 6$

Answer: $\begin{cases} a + b + c = 0 \\ 4a + 2b + c = 3 \\ a - b + c = 6 \end{cases}$ [2 marks]

(b) Solve for $a$, $b$, $c$ [4 marks]

Working: From (1) and (3): $(a+b+c) - (a-b+c) = 0 - 6 = -6$

$2b = -6$, so $b = -3$

From (1): $a + c = 3$

From (3): $a + c = 6 - (-3) = 3$? Check: $a - (-3) + c = 6$, so $a + c + 3 = 6$, thus $a + c = 3$. ✓

From (2): $4a + 2(-3) + c = 3$, so $4a + c = 9$

But $a + c = 3$, so subtract: $3a = 6$, thus $a = 2$

Then $c = 3 - 2 = 1$

Verification: $(2,3)$: $4(2) + 2(-3) + 1 = 8 - 6 + 1 = 3$ ✓

Answer: $a = 2$, $b = -3$, $c = 1$ [4 marks]

(c) Vertex [2 marks]

Working: $y = 2x^2 - 3x + 1$

$x = -\frac{b}{2a} = \frac{3}{4}$

$y = 2(\frac{9}{16}) - 3(\frac{3}{4}) + 1 = \frac{9}{8} - \frac{9}{4} + 1 = \frac{9 - 18 + 8}{8} = -\frac{1}{8}$

Answer: Vertex at $\left(\frac{3}{4}, -\frac{1}{8}\right)$ [2 marks]

Section B Total: 40 marks

Section C: Advanced Coordinate Geometry and Problem Solving (35 marks)

15. Line $y = mx + 2$ tangent to circle $x^2 + y^2 = 4$

(a) Show $16m^2 = 16$ [3 marks]

Working: Substitute: $x^2 + (mx+2)^2 = 4$

$x^2 + m^2x^2 + 4mx + 4 = 4$

$(1+m^2)x^2 + 4mx = 0$

For tangency, discriminant $= 0$:

$B^2 - 4AC = (4m)^2 - 4(1+m^2)(0) = 16m^2 = 0$?

Wait, let me recheck. $C = 0$ in the quadratic. This gives $16m^2 = 0$.

But the problem states $16m^2 = 16$. Let me re-examine...

Actually: $x^2 + y^2 = 4$ is circle centre origin, radius 2.

Line $y = mx + 2$ passes through $(0, 2)$ which is on the circle!

So this line always passes through a point on the circle. For it to be tangent, it should touch at exactly one point. But since $(0,2)$ is on both, we need this to be the only point.

From $(1+m^2)x^2 + 4mx = 0$:

$x((1+m^2)x + 4m) = 0$

So $x = 0$ or $x = -\frac{4m}{1+m^2}$

For single solution (tangency), we need these equal or second to not exist. But $x=0$ always works. For tangency, we need $-\frac{4m}{1+m^2} = 0$, so $m = 0$.

But this gives only $m=0$, not $m = \pm 1$.

Hmm, the problem statement seems inconsistent. Perhaps the intended circle was different, e.g., $x^2 + (y-2)^2 = 4$ or line was $y = mx + c$ with different $c$.

Let me try with line $y = mx + 4$:

$x^2 + (mx+4)^2 = 4$

$(1+m^2)x^2 + 8mx + 16 = 4$

$(1+m^2)x^2 + 8mx + 12 = 0$

Discriminant: $64m^2 - 48(1+m^2) = 64m^2 - 48 - 48m^2 = 16m^2 - 48 = 0$

Gives $m^2 = 3$, not $16m^2 = 16$.

Try line $y = mx + \sqrt{4(1+m^2)}$... this is general tangent condition.

For $x^2+y^2=r^2$, tangent is $y = mx \pm r\sqrt{1+m^2}$. Here requiring $2 = r\sqrt{1+m^2} = 2\sqrt{1+m^2}$, so $1 = 1+m^2$, thus $m=0$.

I think the problem as stated has $y = mx + 2$ and circle $x^2+y^2=4$, but this only gives tangent when $m=0$.

Actually re-reading: the problem says "is tangent to the circle". This implies it is possible. Perhaps I need to re-interpret: maybe the equation doesn't pass through $(0,2)$? No, when $x=0$, $y=2$.

Unless... the circle has different radius? Or perhaps it's $x^2 + y^2 = 2$?

For $x^2+y^2=2$ with $y=mx+2$: $(1+m^2)x^2 + 4mx + 4 = 2$, so $(1+m^2)x^2 + 4mx + 2 = 0$