Secondary 4 Additional Mathematics Practice Paper 1

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

TuitionGoWhere Practice Paper (AI)
Version: 1 of 5
Subject: Additional Mathematics
Level: Secondary 4
Paper: Graphs & Coordinate Geometry Practice
Duration: 1 hour 30 minutes
Total Marks: 80
Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates:

1. Write your Name, Class, and Date in the spaces provided.
2. Answer all questions.
3. Use an approved calculator where appropriate.
4. All necessary working should be clearly shown. Marks may be lost if working is not shown.
5. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

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Section A: Short Answer Questions (30 Marks)

Answer all questions in this section. Each question carries equal marks unless otherwise stated.

1. The line $L_1$ passes through the points $A(2, 5)$ and $B(6, -3)$. Find the equation of the line $L_2$ which is perpendicular to $L_1$ and passes through the midpoint of $AB$. [3]

2. Find the coordinates of the points of intersection of the curve $y = x^2 - 4x + 3$ and the line $y = x - 1$. [3]

3. The circle $C$ has equation $x^2 + y^2 - 6x + 4y - 12 = 0$. Find the coordinates of the centre and the radius of the circle. [3]

4. Determine the set of values of $k$ for which the line $y = kx + 2$ does not intersect the curve $y = x^2 - 3x + 5$. [3]

5. Points $P(1, 2)$, $Q(5, 6)$, and $R(9, 2)$ are vertices of a triangle. Show that triangle $PQR$ is isosceles and find its area. [4]

6. The point $A$ has coordinates $(3, -1)$ and the point $B$ has coordinates $(7, 5)$. Find the equation of the perpendicular bisector of $AB$ in the form $ax + by + c = 0$, where $a, b, c$ are integers. [4]

7. A curve has equation $y = \frac{12}{x}$. The tangent to the curve at the point where $x = 3$ intersects the x-axis at point $A$ and the y-axis at point $B$. Find the coordinates of $A$ and $B$. [4]

8. Find the equation of the circle which passes through the origin $O(0,0)$ and has its centre at $(4, -3)$. [3]

9. The lines $y = 2x + 1$ and $y = -x + 7$ intersect at point $P$. Find the distance of point $P$ from the origin. [3]

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Section B: Structured Questions (30 Marks)

Answer all questions in this section.

10. The diagram shows a triangle $ABC$ with vertices $A(1, 1)$, $B(5, 3)$, and $C(3, 7)$.

<image_placeholder> id: Q10-fig1 type: diagram linked_question: Q10 description: A triangle ABC plotted on a Cartesian plane. Vertex A is at (1,1), B is at (5,3), and C is at (3,7). The axes are labeled x and y. Grid lines are visible. labels: A(1,1), B(5,3), C(3,7), O(0,0) values: Coordinates as specified. must_show: The triangle shape, vertices labeled, and the coordinate axes. </image_placeholder>

(a) Find the gradient of the line $AC$. [1]
(b) Find the equation of the altitude from $B$ to $AC$. [3]
(c) Find the coordinates of the foot of the perpendicular from $B$ to $AC$. [3]
(d) Hence, or otherwise, calculate the area of triangle $ABC$. [3]

11. The curve $C_1$ has equation $y = x^2 - 2x - 3$ and the curve $C_2$ has equation $y = -x^2 + 4x + 1$.

(a) Find the coordinates of the points of intersection of $C_1$ and $C_2$. [4]
(b) Find the equation of the line passing through these two points of intersection. [2]
(c) Determine whether the line found in part (b) is parallel to the line $y = 3x$. Give a reason for your answer. [1]

12. A circle has centre $C(2, 1)$ and radius $5$.

(a) Write down the equation of the circle. [1]
(b) The line $L$ has equation $y = 2x + k$. Find the values of $k$ for which the line $L$ is a tangent to the circle. [5]
(c) For the case where $k > 0$, find the coordinates of the point of contact between the line and the circle. [4]

13. The points $A(-2, 0)$ and $B(4, 6)$ lie on a circle. The centre of the circle lies on the line $y = x$.

(a) Find the equation of the perpendicular bisector of $AB$. [4]
(b) Find the coordinates of the centre of the circle. [2]
(c) Find the equation of the circle. [2]
(d) Determine whether the point $D(6, 2)$ lies inside, on, or outside the circle. Show your working. [2]

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Section C: Problem Solving (20 Marks)

Answer all questions in this section.

14. The diagram shows a rectangle $ABCD$. The vertex $A$ is at $(1, 2)$ and the vertex $C$ is at $(7, 8)$. The side $AB$ lies on the line with equation $y = 2x$.

<image_placeholder> id: Q14-fig1 type: diagram linked_question: Q14 description: A rectangle ABCD plotted on a Cartesian plane. Vertex A is at (1,2) and C is at (7,8). The line containing side AB passes through the origin and A. The rectangle is tilted. labels: A(1,2), C(7,8), Line y=2x values: Coordinates as specified. must_show: Rectangle ABCD, diagonal AC, line y=2x extending through A and B. </image_placeholder>

(a) Find the equation of the line $BC$. [3]
(b) Find the coordinates of vertex $B$. [4]
(c) Find the coordinates of vertex $D$. [2]
(d) Calculate the area of rectangle $ABCD$. [3]

15. Two circles $C_1$ and $C_2$ have equations: $C_1: x^2 + y^2 - 4x - 6y - 12 = 0$ $C_2: x^2 + y^2 + 2x - 8y + 13 = 0$

(a) Find the coordinates of the centres and the radii of $C_1$ and $C_2$. [4]
(b) Show that the two circles intersect at two distinct points. [3]
(c) Find the equation of the common chord of the two circles. [3]
(d) Find the length of the common chord. [4]

16. The point $P$ moves such that its distance from the point $A(0, 4)$ is always twice its distance from the point $B(3, 0)$.

(a) Find the equation of the locus of $P$. [5]
(b) Identify the shape of the locus and state its centre and radius. [3]
(c) Determine whether the origin lies inside or outside this locus. [2]

17. The line $y = mx + c$ is a tangent to the circle $x^2 + y^2 = 25$.

(a) Show that $c^2 = 25(1 + m^2)$. [4]
(b) Given that the tangent passes through the point $(7, 1)$, find the possible values of $m$. [4]
(c) Hence, find the equations of the two tangents from $(7, 1)$ to the circle. [2]

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Answer Key - Additional Mathematics Secondary 4

Topic: Graphs & Coordinate Geometry
Version: 1 of 5

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Section A: Short Answer Questions

1.
Step 1: Find gradient of $L_1$.
$m_{L1} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2$.
Step 2: Find gradient of $L_2$ (perpendicular).
$m_{L2} = -\frac{1}{m_{L1}} = -\frac{1}{-2} = \frac{1}{2}$.
Step 3: Find midpoint of $AB$.
$M = \left(\frac{2+6}{2}, \frac{5+(-3)}{2}\right) = (4, 1)$.
Step 4: Equation of $L_2$.
$y - 1 = \frac{1}{2}(x - 4) \Rightarrow 2(y - 1) = x - 4 \Rightarrow 2y - 2 = x - 4$.
$x - 2y - 2 = 0$ or $y = \frac{1}{2}x + 1$.
Answer: $x - 2y - 2 = 0$ [3]

2.
Step 1: Equate $y$ values.
$x^2 - 4x + 3 = x - 1$
$x^2 - 5x + 4 = 0$
Step 2: Solve for $x$.
$(x - 4)(x - 1) = 0 \Rightarrow x = 1$ or $x = 4$.
Step 3: Find corresponding $y$.
If $x = 1, y = 1 - 1 = 0$. Point $(1, 0)$.
If $x = 4, y = 4 - 1 = 3$. Point $(4, 3)$.
Answer: $(1, 0)$ and $(4, 3)$ [3]

3.
Step 1: Complete the square for $x$ and $y$.
$x^2 - 6x + y^2 + 4y = 12$
$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$
$(x - 3)^2 + (y + 2)^2 = 25$
Step 2: Identify centre and radius.
Centre $(3, -2)$, Radius $\sqrt{25} = 5$.
Answer: Centre $(3, -2)$, Radius $5$ [3]

4.
Step 1: Set up intersection equation.
$kx + 2 = x^2 - 3x + 5$
$x^2 - (3 + k)x + 3 = 0$
Step 2: Condition for no intersection is discriminant $< 0$.
$b^2 - 4ac < 0$
$(-(3+k))^2 - 4(1)(3) < 0$
$(3+k)^2 - 12 < 0$
Step 3: Solve inequality.
$(3+k)^2 < 12$
$-\sqrt{12} < 3+k < \sqrt{12}$
$-2\sqrt{3} - 3 < k < 2\sqrt{3} - 3$
Answer: $-3 - 2\sqrt{3} < k < -3 + 2\sqrt{3}$ [3]

5.
Step 1: Calculate side lengths.
$PQ = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16+16} = \sqrt{32}$
$QR = \sqrt{(9-5)^2 + (2-6)^2} = \sqrt{16+16} = \sqrt{32}$
$PR = \sqrt{(9-1)^2 + (2-2)^2} = \sqrt{64} = 8$
Since $PQ = QR$, it is isosceles.
Step 2: Find area.
Base $PR$ is horizontal, length $8$.
Height is vertical distance from $Q(5,6)$ to line $y=2$ (line PR). Height $= 6 - 2 = 4$.
Area $= \frac{1}{2} \times 8 \times 4 = 16$.
Answer: Isosceles shown, Area $= 16$ [4]

6.
Step 1: Midpoint of $AB$.
$M = (\frac{3+7}{2}, \frac{-1+5}{2}) = (5, 2)$.
Step 2: Gradient of $AB$.
$m_{AB} = \frac{5 - (-1)}{7 - 3} = \frac{6}{4} = \frac{3}{2}$.
Step 3: Gradient of perpendicular bisector.
$m_{\perp} = -\frac{2}{3}$.
Step 4: Equation.
$y - 2 = -\frac{2}{3}(x - 5)$
$3(y - 2) = -2(x - 5)$
$3y - 6 = -2x + 10$
$2x + 3y - 16 = 0$.
Answer: $2x + 3y - 16 = 0$ [4]

7.
Step 1: Find point on curve.
$x = 3 \Rightarrow y = \frac{12}{3} = 4$. Point $(3, 4)$.
Step 2: Find gradient of tangent.
$y = 12x^{-1} \Rightarrow \frac{dy}{dx} = -12x^{-2} = -\frac{12}{x^2}$.
At $x = 3, m = -\frac{12}{9} = -\frac{4}{3}$.
Step 3: Equation of tangent.
$y - 4 = -\frac{4}{3}(x - 3)$
$3(y - 4) = -4(x - 3)$
$3y - 12 = -4x + 12$
$4x + 3y = 24$.
Step 4: Find intercepts.
x-intercept ($y=0$): $4x = 24 \Rightarrow x = 6$. $A(6, 0)$.
y-intercept ($x=0$): $3y = 24 \Rightarrow y = 8$. $B(0, 8)$.
Answer: $A(6, 0), B(0, 8)$ [4]

8.
Step 1: Radius is distance from centre $(4, -3)$ to origin $(0,0)$.
$r^2 = (4-0)^2 + (-3-0)^2 = 16 + 9 = 25$.
Step 2: Equation.
$(x - 4)^2 + (y + 3)^2 = 25$
$x^2 - 8x + 16 + y^2 + 6y + 9 = 25$
$x^2 + y^2 - 8x + 6y = 0$.
Answer: $x^2 + y^2 - 8x + 6y = 0$ [3]

9.
Step 1: Find intersection $P$.
$2x + 1 = -x + 7 \Rightarrow 3x = 6 \Rightarrow x = 2$.
$y = 2(2) + 1 = 5$. $P(2, 5)$.
Step 2: Distance from origin.
$OP = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$.
Answer: $\sqrt{29}$ [3]

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Section B: Structured Questions

10.
(a) Gradient $AC = \frac{7 - 1}{3 - 1} = \frac{6}{2} = 3$. [1]
(b) Gradient of altitude from $B$ is $-\frac{1}{3}$.
Passes through $B(5, 3)$.
$y - 3 = -\frac{1}{3}(x - 5) \Rightarrow 3(y - 3) = -(x - 5) \Rightarrow 3y - 9 = -x + 5$.
$x + 3y - 14 = 0$. [3]
(c) Solve simultaneous equations for foot of perpendicular ($F$).
Line $AC$: $y - 1 = 3(x - 1) \Rightarrow y = 3x - 2$.
Substitute into altitude eq: $x + 3(3x - 2) - 14 = 0 \Rightarrow x + 9x - 6 - 14 = 0 \Rightarrow 10x = 20 \Rightarrow x = 2$.
$y = 3(2) - 2 = 4$.
Foot is $(2, 4)$. [3]
(d) Base $AC = \sqrt{(3-1)^2 + (7-1)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$.
Height $BF = \sqrt{(5-2)^2 + (3-4)^2} = \sqrt{9 + 1} = \sqrt{10}$.
Area $= \frac{1}{2} \times 2\sqrt{10} \times \sqrt{10} = 10$. [3]
(Alternative: Shoelace formula or box method yields same result)

11.
(a) $x^2 - 2x - 3 = -x^2 + 4x + 1 \Rightarrow 2x^2 - 6x - 4 = 0 \Rightarrow x^2 - 3x - 2 = 0$.
$x = \frac{3 \pm \sqrt{9 - 4(1)(-2)}}{2} = \frac{3 \pm \sqrt{17}}{2}$.
Let $x_1 = \frac{3 + \sqrt{17}}{2}, x_2 = \frac{3 - \sqrt{17}}{2}$.
$y_1 = x_1 - 1$? No, use linear eq from subtraction?
Actually, subtracting the two curve equations gives the line through intersections directly (see part b).
Let's find y using $y = x^2 - 2x - 3$.
This is messy. Better to find the line first? No, question asks for coordinates.
$y = \frac{3 \pm \sqrt{17}}{2} - 1$? Wait, is $y=x-1$ the line?
Subtracting $C_1$ from $C_2$: $2x^2 - 6x - 4 = 0$ is not a line.
Wait, $C_1: y = x^2 - 2x - 3$, $C_2: y = -x^2 + 4x + 1$.
Intersection: $2x^2 - 6x - 4 = 0 \implies x^2 - 3x - 2 = 0$.
Roots are irrational.
$y = x^2 - 2x - 3$. Since $x^2 = 3x + 2$, $y = (3x + 2) - 2x - 3 = x - 1$.
So $y_1 = x_1 - 1 = \frac{3 + \sqrt{17}}{2} - \frac{2}{2} = \frac{1 + \sqrt{17}}{2}$.
$y_2 = x_2 - 1 = \frac{3 - \sqrt{17}}{2} - \frac{2}{2} = \frac{1 - \sqrt{17}}{2}$.
Points: $\left(\frac{3 + \sqrt{17}}{2}, \frac{1 + \sqrt{17}}{2}\right)$ and $\left(\frac{3 - \sqrt{17}}{2}, \frac{1 - \sqrt{17}}{2}\right)$. [4]
(b) The line passing through intersections is found by subtracting the equations?
Actually, we found $y = x - 1$ during substitution.
Equation: $y = x - 1$ or $x - y - 1 = 0$. [2]
(c) Gradient of $y = x - 1$ is $1$. Gradient of $y = 3x$ is $3$.
$1 \neq 3$, so not parallel. [1]

12.
(a) $(x - 2)^2 + (y - 1)^2 = 25$. [1]
(b) Substitute $y = 2x + k$ into circle eq.
$(x - 2)^2 + (2x + k - 1)^2 = 25$
$x^2 - 4x + 4 + 4x^2 + 4x(k - 1) + (k - 1)^2 = 25$
$5x^2 + x[-4 + 4k - 4] + [4 + k^2 - 2k + 1 - 25] = 0$
$5x^2 + (4k - 8)x + (k^2 - 2k - 20) = 0$.
For tangent, discriminant $= 0$.
$(4k - 8)^2 - 4(5)(k^2 - 2k - 20) = 0$
$16(k - 2)^2 - 20(k^2 - 2k - 20) = 0$
Divide by 4: $4(k^2 - 4k + 4) - 5(k^2 - 2k - 20) = 0$
$4k^2 - 16k + 16 - 5k^2 + 10k + 100 = 0$
$-k^2 - 6k + 116 = 0 \Rightarrow k^2 + 6k - 116 = 0$.
$k = \frac{-6 \pm \sqrt{36 - 4(1)(-116)}}{2} = \frac{-6 \pm \sqrt{36 + 464}}{2} = \frac{-6 \pm \sqrt{500}}{2} = \frac{-6 \pm 10\sqrt{5}}{2} = -3 \pm 5\sqrt{5}$. [5]
(c) $k > 0 \Rightarrow k = -3 + 5\sqrt{5}$.
Solve for $x$ using $x = \frac{-b}{2a}$ from quadratic formula (since disc=0).
$x = \frac{-(4k - 8)}{10} = \frac{8 - 4k}{10} = \frac{4 - 2k}{5}$.
Substitute $k$: $x = \frac{4 - 2(-3 + 5\sqrt{5})}{5} = \frac{4 + 6 - 10\sqrt{5}}{5} = \frac{10 - 10\sqrt{5}}{5} = 2 - 2\sqrt{5}$.
$y = 2x + k = 2(2 - 2\sqrt{5}) + (-3 + 5\sqrt{5}) = 4 - 4\sqrt{5} - 3 + 5\sqrt{5} = 1 + \sqrt{5}$.
Point of contact: $(2 - 2\sqrt{5}, 1 + \sqrt{5})$. [4]

13.
(a) Midpoint of $AB$: $(\frac{-2+4}{2}, \frac{0+6}{2}) = (1, 3)$.
Gradient $AB$: $\frac{6-0}{4-(-2)} = \frac{6}{6} = 1$.
Gradient perp bisector: $-1$.
Eq: $y - 3 = -1(x - 1) \Rightarrow y = -x + 4$. [4]
(b) Centre lies on $y = x$ and $y = -x + 4$.
$x = -x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$.
$y = 2$. Centre $(2, 2)$. [2]
(c) Radius squared $r^2 = (2 - (-2))^2 + (2 - 0)^2 = 4^2 + 2^2 = 20$.
Eq: $(x - 2)^2 + (y - 2)^2 = 20$ or $x^2 + y^2 - 4x - 4y - 12 = 0$. [2]
(d) Distance $CD^2 = (6 - 2)^2 + (2 - 2)^2 = 16 + 0 = 16$.
$r^2 = 20$. Since $16 < 20$, $D$ is inside the circle. [2]

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Section C: Problem Solving

14.
(a) Gradient $AB = 2$. Since $ABCD$ is a rectangle, $BC \perp AB$.
Gradient $BC = -\frac{1}{2}$.
Passes through $C(7, 8)$.
$y - 8 = -\frac{1}{2}(x - 7) \Rightarrow 2(y - 8) = -(x - 7) \Rightarrow 2y - 16 = -x + 7$.
$x + 2y - 23 = 0$. [3]
(b) $B$ is intersection of $y = 2x$ and $x + 2y - 23 = 0$.
$x + 2(2x) - 23 = 0 \Rightarrow 5x = 23 \Rightarrow x = 4.6$.
$y = 2(4.6) = 9.2$.
$B(4.6, 9.2)$. [4]
(c) Midpoint of $AC$ is same as midpoint of $BD$.
$M_{AC} = (\frac{1+7}{2}, \frac{2+8}{2}) = (4, 5)$.
Let $D(x, y)$. $\frac{x + 4.6}{2} = 4 \Rightarrow x + 4.6 = 8 \Rightarrow x = 3.4$.
$\frac{y + 9.2}{2} = 5 \Rightarrow y + 9.2 = 10 \Rightarrow y = 0.8$.
$D(3.4, 0.8)$. [2]
(d) Length $AB = \sqrt{(4.6 - 1)^2 + (9.2 - 2)^2} = \sqrt{3.6^2 + 7.2^2} = \sqrt{12.96 + 51.84} = \sqrt{64.8}$.
Length $BC = \sqrt{(7 - 4.6)^2 + (8 - 9.2)^2} = \sqrt{2.4^2 + (-1.2)^2} = \sqrt{5.76 + 1.44} = \sqrt{7.2}$.
Area $= \sqrt{64.8} \times \sqrt{7.2} = \sqrt{466.56} = 21.6$. [3]

15.
(a) $C_1: (x - 2)^2 - 4 + (y - 3)^2 - 9 - 12 = 0 \Rightarrow (x - 2)^2 + (y - 3)^2 = 25$.
Centre $O_1(2, 3)$, $r_1 = 5$.
$C_2: (x + 1)^2 - 1 + (y - 4)^2 - 16 + 13 = 0 \Rightarrow (x + 1)^2 + (y - 4)^2 = 4$.
Centre $O_2(-1, 4)$, $r_2 = 2$. [4]
(b) Distance $O_1O_2 = \sqrt{(-1 - 2)^2 + (4 - 3)^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.16$.
Sum of radii $= 5 + 2 = 7$. Diff of radii $= 5 - 2 = 3$.
Since $3 < \sqrt{10} < 7$, the circles intersect at two distinct points. [3]
(c) Subtract equations:
$(x^2 + y^2 - 4x - 6y - 12) - (x^2 + y^2 + 2x - 8y + 13) = 0$
$-6x + 2y - 25 = 0 \Rightarrow 6x - 2y + 25 = 0$. [3]
(d) Distance from $O_1(2, 3)$ to line $6x - 2y + 25 = 0$.
$d = \frac{|6(2) - 2(3) + 25|}{\sqrt{6^2 + (-2)^2}} = \frac{|12 - 6 + 25|}{\sqrt{40}} = \frac{31}{\sqrt{40}}$.
Half-chord length $h = \sqrt{r_1^2 - d^2} = \sqrt{25 - \frac{961}{40}} = \sqrt{\frac{1000 - 961}{40}} = \sqrt{\frac{39}{40}}$.
Total length $= 2h = 2\sqrt{\frac{39}{40}} = \sqrt{\frac{156}{40}} = \sqrt{3.9} \approx 1.97$. [4]

16.
(a) Let $P(x, y)$. $PA = 2 PB$.
$PA^2 = 4 PB^2$.
$x^2 + (y - 4)^2 = 4[(x - 3)^2 + y^2]$.
$x^2 + y^2 - 8y + 16 = 4[x^2 - 6x + 9 + y^2]$.
$x^2 + y^2 - 8y + 16 = 4x^2 - 24x + 36 + 4y^2$.
$3x^2 + 3y^2 - 24x + 8y + 20 = 0$.
$x^2 + y^2 - 8x + \frac{8}{3}y + \frac{20}{3} = 0$. [5]
(b) Complete square:
$(x - 4)^2 - 16 + (y + \frac{4}{3})^2 - \frac{16}{9} + \frac{20}{3} = 0$.
$(x - 4)^2 + (y + \frac{4}{3})^2 = 16 + \frac{16}{9} - \frac{60}{9} = 16 - \frac{44}{9} = \frac{144 - 44}{9} = \frac{100}{9}$.
Circle, Centre $(4, -\frac{4}{3})$, Radius $\frac{10}{3}$. [3]
(c) Distance from Origin to Centre $(4, -4/3)$:
$d^2 = 16 + \frac{16}{9} = \frac{160}{9}$.
Radius squared $r^2 = \frac{100}{9}$.
Since $\frac{160}{9} > \frac{100}{9}$, Origin is outside. [2]

17.
(a) Substitute $y = mx + c$ into $x^2 + y^2 = 25$.
$x^2 + (mx + c)^2 = 25 \Rightarrow (1 + m^2)x^2 + 2mcx + c^2 - 25 = 0$.
Tangent $\Rightarrow$ Discriminant $= 0$.
$(2mc)^2 - 4(1 + m^2)(c^2 - 25) = 0$.
$4m^2c^2 - 4(c^2 - 25 + m^2c^2 - 25m^2) = 0$.
$m^2c^2 - c^2 + 25 - m^2c^2 + 25m^2 = 0$.
$-c^2 + 25 + 25m^2 = 0 \Rightarrow c^2 = 25(1 + m^2)$. [4]
(b) Line passes through $(7, 1) \Rightarrow 1 = 7m + c \Rightarrow c = 1 - 7m$.
Substitute into (a):
$(1 - 7m)^2 = 25(1 + m^2)$.
$1 - 14m + 49m^2 = 25 + 25m^2$.
$24m^2 - 14m - 24 = 0$.
$12m^2 - 7m - 12 = 0$.
$(4m + 3)(3m - 4) = 0$.
$m = -\frac{3}{4}$ or $m = \frac{4}{3}$. [4]
(c) If $m = -\frac{3}{4}, c = 1 - 7(-\frac{3}{4}) = 1 + \frac{21}{4} = \frac{25}{4}$.
Eq: $y = -\frac{3}{4}x + \frac{25}{4} \Rightarrow 3x + 4y - 25 = 0$.
If $m = \frac{4}{3}, c = 1 - 7(\frac{4}{3}) = 1 - \frac{28}{3} = -\frac{25}{3}$.
Eq: $y = \frac{4}{3}x - \frac{25}{3} \Rightarrow 4x - 3y - 25 = 0$. [2]