Geometries — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for geometries.

UT-1: Circle Theorems — Angle at Centre

Question:

In the figure, $A$, $B$, $C$ are points on a circle with centre $O$. If $\angle ABC = 35°$ and $\angle BAC = 70°$, find $\angle AOC$.

Solution:

The angle at the centre is twice the angle at the circumference subtended by the same arc.

$\angle ABC = 35°$ is subtended by arc $AC$.

$\angle AOC = 2 \times \angle ABC = 2 \times 35° = 70°$.

Alternatively, in triangle $ABC$: $\angle ACB = 180° - 35° - 70° = 75°$.

$\angle AOC$ subtended by arc $ABC$ (the major arc) $= 2 \times 75° = 150°$.

Since $\angle AOC$ on the minor arc $AC$ plus the reflex angle $= 360°$:

Minor $\angle AOC = 2 \times 35° = 70°$.

A common mistake is confusing which arc the angle subtends. $\angle ABC$ subtends arc $AC$ (not containing $B$), so the angle at the centre is $70°$.

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UT-2: Tangent-Radius Perpendicularity

Question:

$TA$ and $TB$ are tangents to a circle with centre $O$, touching the circle at $A$ and $B$ respectively. If $\angle ATB = 50°$, find $\angle AOB$.

Solution:

$OA \perp TA$ and $OB \perp TB$ (tangent perpendicular to radius).

In quadrilateral $OATB$: $\angle OAT = \angle OBT = 90°$.

$\angle AOB = 360° - 90° - 90° - 50° = 130°$

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UT-3: Cyclic Quadrilateral Properties

Question:

$ABCD$ is a cyclic quadrilateral with $\angle A = 110°$ and $\angle C = 2\angle B$. Find $\angle B$ and $\angle D$.

Solution:

In a cyclic quadrilateral, opposite angles sum to $180°$:

$\angle A + \angle C = 180° \implies 110° + 2\angle B = 180° \implies 2\angle B = 70° \implies \angle B = 35°$.

$\angle B + \angle D = 180° \implies \angle D = 145°$.

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UT-4: Vector Geometry — Collinearity

Question:

Let $\vec{OA} = \mathbf{'\{'}a{'\}'}$, $\vec{OB} = \mathbf{'\{'}b{'\}'}$. Point $P$ divides $AB$ in the ratio $2:1$ and point $Q$ divides $AB$ in the ratio $3:2$. Express $\vec{PQ}$ in terms of $\mathbf{'\{'}a{'\}'}$ and $\mathbf{'\{'}b{'\}'}$.

Solution:

$\vec{AB} = \mathbf{'\{'}b{'\}'} - \mathbf{'\{'}a{'\}'}$.

$P$ divides $AB$ in ratio $2:1$ (i.e. $AP:PB = 2:1$):

$\vec{OP} = \vec{OA} + \frac{2}{3}\vec{AB} = \mathbf{'\{'}a{'\}'} + \frac{2}{3}(\mathbf{'\{'}b{'\}'} - \mathbf{'\{'}a{'\}'}) = \frac{1}{3}\mathbf{'\{'}a{'\}'} + \frac{2}{3}\mathbf{'\{'}b{'\}'}$

$Q$ divides $AB$ in ratio $3:2$ (i.e. $AQ:QB = 3:2$):

$\vec{OQ} = \vec{OA} + \frac{3}{5}\vec{AB} = \mathbf{'\{'}a{'\}'} + \frac{3}{5}(\mathbf{'\{'}b{'\}'} - \mathbf{'\{'}a{'\}'}) = \frac{2}{5}\mathbf{'\{'}a{'\}'} + \frac{3}{5}\mathbf{'\{'}b{'\}'}$

$\vec{PQ} = \vec{OQ} - \vec{OP} = \left(\frac{2}{5} - \frac{1}{3}\right)\mathbf{'\{'}a{'\}'} + \left(\frac{3}{5} - \frac{2}{3}\right)\mathbf{'\{'}b{'\}'} = \frac{1}{15}\mathbf{'\{'}a{'\}'} - \frac{1}{15}\mathbf{'\{'}b{'\}'} = \frac{1}{15}(\mathbf{'\{'}a{'\}'} - \mathbf{'\{'}b{'\}'})$

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UT-5: Coordinate Proof Using Algebra

Question:

Use coordinate geometry to prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from all three vertices.

Solution:

Place the right angle at the origin, with the legs along the axes.

Let $A = (0, 0)$, $B = (2a, 0)$, $C = (0, 2b)$.

Right angle at $A$, hypotenuse is $BC$.

Midpoint $M$ of $BC$: $\left(\dfrac{2a + 0}{2},\; \dfrac{0 + 2b}{2}\right) = (a,\; b)$.

$MA = \sqrt{a^2 + b^2}$.

$MB = \sqrt{(a - 2a)^2 + (b - 0)^2} = \sqrt{a^2 + b^2}$.

$MC = \sqrt{(a - 0)^2 + (b - 2b)^2} = \sqrt{a^2 + b^2}$.

Since $MA = MB = MC$, the midpoint of the hypotenuse is equidistant from all three vertices.

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Integration Tests

Tests synthesis of geometries with other topics.

IT-1: Geometries and Trigonometry (with Trigonometry)

Question:

In triangle $ABC$, $AB = 10$ cm, $BC = 8$ cm, $CA = 6$ cm. Find the radius of the circumscribed circle.

Solution:

First check it is a right triangle: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$. Yes, right angle at $C$.

In a right triangle, the hypotenuse is the diameter of the circumscribed circle.

Radius $= \dfrac{AB}{2} = \dfrac{10}{2} = 5$ cm.

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IT-2: Geometries and Vectors (with Coordinate Geometry)

Question:

The position vectors of $A$ and $B$ are $\mathbf{'\{'}a{'\}'} = 2\mathbf{'\{'}i{'\}'} + 3\mathbf{'\{'}j{'\}'}$ and $\mathbf{'\{'}b{'\}'} = 8\mathbf{'\{'}i{'\}'} - \mathbf{'\{'}j{'\}'}$. Point $C$ has position vector $\mathbf{'\{'}c{'\}'} = 4\mathbf{'\{'}i{'\}'} + 7\mathbf{'\{'}j{'\}'}$. Prove that $A$, $B$, $C$ form an isosceles triangle.

Solution:

$\vec{AB} = \mathbf{'\{'}b{'\}'} - \mathbf{'\{'}a{'\}'} = 6\mathbf{'\{'}i{'\}'} - 4\mathbf{'\{'}j{'\}'}$

$|\vec{AB}| = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$

$\vec{AC} = \mathbf{'\{'}c{'\}'} - \mathbf{'\{'}a{'\}'} = 2\mathbf{'\{'}i{'\}'} + 4\mathbf{'\{'}j{'\}'}$

$|\vec{AC}| = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$

$\vec{BC} = \mathbf{'\{'}c{'\}'} - \mathbf{'\{'}b{'\}'} = -4\mathbf{'\{'}i{'\}'} + 8\mathbf{'\{'}j{'\}'}$

$|\vec{BC}| = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$

$|\vec{AB}| = 2\sqrt{13}$, $|\vec{AC}| = 2\sqrt{5}$, $|\vec{BC}| = 4\sqrt{5}$.

None are equal, so this is not isosceles. Verifying: $|\vec{AC}|^2 + |\vec{BC}|^2 = 20 + 80 = 100 \neq 52 = |\vec{AB}|^2$, confirming it is also not right-angled. The triangle is scalene.

Key takeaway: Always compute and verify rather than assuming geometric properties.

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IT-3: Geometries and Algebra (with Polynomials)

Question:

The points $(1, 2)$, $(3, 6)$, and $(k, 10)$ are collinear. Find $k$.

Solution:

Collinearity means equal slopes:

$\frac{6 - 2}{3 - 1} = \frac{10 - 6}{k - 3}$

$\frac{4}{2} = \frac{4}{k - 3}$

$2 = \frac{4}{k - 3}$

$k - 3 = 2 \implies k = 5$

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Worked Examples

WE-1: Angle Between Tangent and Chord

Question:

$TA$ is a tangent to a circle at $A$. $AB$ is a chord of the circle. If $\angle BAT = 42°$, find the angle in the alternate segment, i.e. the angle subtended by chord $AB$ in the opposite segment.

Solution:

The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.

Therefore the angle in the alternate segment $= 42°$.

If $C$ is any point on the circle on the opposite side of $AB$ from $T$, then $\angle ACB = 42°$.

DSE Exam Technique: When stating circle theorems, name the theorem explicitly. The HKEAA requires you to cite the specific theorem being used, e.g. "By the alternate segment theorem."

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WE-2: Chord Length from Central Angle

Question:

A circle has radius $r = 10$ cm. A chord subtends a central angle of $120°$. Find the length of the chord.

Solution:

Let the chord be $AB$ with centre $O$. Then $\angle AOB = 120°$ and $OA = OB = 10$ cm.

Drop the perpendicular from $O$ to $AB$, meeting at $M$.

Since $OM$ bisects $\angle AOB$: $\angle AOM = 60°$.

$AM = OA \sin 60° = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$

$AB = 2 \times AM = 10\sqrt{3} \text{ cm}$

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WE-3: Proving Cyclic Quadrilateral

Question:

In triangle $ABC$, $D$ is a point on $BC$ such that $AD$ bisects $\angle BAC$. If $AD = AB$, prove that $ABCD$ is a cyclic quadrilateral.

Solution:

Since $AD = AB$, triangle $ABD$ is isosceles with $\angle ABD = \angle ADB$. ... (1)

Since $AD$ bisects $\angle BAC$: $\angle BAD = \angle CAD$. ... (2)

The exterior angle of triangle $ABD$ at $D$: $\angle ADB + \angle BAD + \angle ABD = 180°$.

In triangle $ADC$: $\angle CAD + \angle ADC + \angle ACD = 180°$.

Using $\angle ADB = \angle ABD$ from (1) and $\angle BAD = \angle CAD$ from (2):

$\angle ABD + \angle CAD + \angle ABD = 180°$ and $\angle CAD + \angle ADC + \angle ACD = 180°$.

Also $\angle ADB + \angle ADC = 180°$ (angles on a straight line), so $\angle ABD + \angle ADC = 180°$.

Since $\angle ABD + \angle ACD = 180°$ (from the two triangle angle sums), we have $\angle ADC = \angle ACD$, meaning... Let us reconsider.

From the isosceles triangle: $\angle ABD = \angle ADB$.

In triangle $ABC$: $\angle ABC = \angle ABD = \angle ADB$.

Since $\angle ADB$ and $\angle ADC$ are supplementary (straight line): $\angle ADB + \angle ADC = 180°$.

So $\angle ABC + \angle ADC = 180°$.

Therefore $ABCD$ is a cyclic quadrilateral (opposite angles are supplementary).

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WE-4: Intersecting Chords

Question:

Two chords $AB$ and $CD$ of a circle intersect at $P$ inside the circle. Given that $AP = 4$ cm, $PB = 6$ cm, and $CP = 3$ cm, find $PD$.

Solution:

By the intersecting chords theorem: $AP \times PB = CP \times PD$.

$4 \times 6 = 3 \times PD$

$PD = \frac{24}{3} = 8 \text{ cm}$

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WE-5: Area of Sector and Segment

Question:

A sector of a circle with radius $8$ cm and central angle $135°$ is drawn. Find:

(a) The area of the sector. (b) The area of the segment (the region between the chord and the arc).

Solution:

(a) Area of sector $= \dfrac{\theta}{360°} \times \pi r^2 = \dfrac{135°}{360°} \times \pi \times 64 = \dfrac{3}{8} \times 64\pi = 24\pi$ cm$^2$.

(b) The triangle formed by the two radii and the chord:

$\text{Area of triangle} = \frac{1}{2} \times 8 \times 8 \times \sin 135° = 32 \times \frac{\sqrt{2}}{2} = 16\sqrt{2} \text{ cm}^2$

$\text{Area of segment} = 24\pi - 16\sqrt{2} \text{ cm}^2$

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WE-6: Vector Proof of Midpoint

Question:

In triangle $ABC$, let $D$ be the midpoint of $BC$. Using vectors, prove that $\vec{AD} = \dfrac{1}{2}(\vec{AB} + \vec{AC})$.

Solution:

$\vec{AD} = \vec{AB} + \vec{BD}$

Since $D$ is the midpoint of $BC$: $\vec{BD} = \dfrac{1}{2}\vec{BC}$.

$\vec{BC} = \vec{BA} + \vec{AC} = -\vec{AB} + \vec{AC}$

Therefore:

$\vec{AD} = \vec{AB} + \frac{1}{2}(-\vec{AB} + \vec{AC}) = \vec{AB} - \frac{1}{2}\vec{AB} + \frac{1}{2}\vec{AC} = \frac{1}{2}\vec{AB} + \frac{1}{2}\vec{AC} = \frac{1}{2}(\vec{AB} + \vec{AC})$

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WE-7: Finding the Centre of a Circle Through Three Points

Question:

Find the equation of the circle passing through the points $(0, 0)$, $(4, 0)$, and $(0, 6)$.

Solution:

Let the circle have equation $x^2 + y^2 + Dx + Ey + F = 0$.

Substituting $(0, 0)$: $F = 0$.

Substituting $(4, 0)$: $16 + 4D = 0 \implies D = -4$.

Substituting $(0, 6)$: $36 + 6E = 0 \implies E = -6$.

Equation: $x^2 + y^2 - 4x - 6y = 0$.

Completing the square: $(x - 2)^2 + (y - 3)^2 = 4 + 9 = 13$.

Centre: $(2, 3)$, Radius: $\sqrt{13}$.

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WE-8: Parallel Lines and Transversal Angles

Question:

In the figure, $AB \parallel CD$. $EF$ is a transversal cutting $AB$ at $G$ and $CD$ at $H$. If $\angle EGB = 3x + 10°$ and $\angle CHG = 5x - 30°$, find $x$.

Solution:

Since $AB \parallel CD$ and $EF$ is a transversal, $\angle EGB$ and $\angle CHG$ are supplementary (interior angles on the same side of the transversal).

$\angle EGB + \angle CHG = 180°$

$(3x + 10°) + (5x - 30°) = 180°$

$8x - 20° = 180°$

$8x = 200°$

$x = 25°$

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Common Pitfalls

1. Confusing the angle at the centre with the angle at the circumference. The angle at the centre is twice the angle at the circumference, but only when they subtend the same arc. Identifying the correct arc is essential. A common DSE error is using the wrong arc.

2. Assuming the tangent is perpendicular to the chord. The tangent is perpendicular to the radius at the point of contact, not to the chord. The perpendicular from the centre to a chord bisects the chord, but this is a different property.

3. Forgetting that cyclic quadrilateral conditions work both ways. If opposite angles sum to $180°$, then the quadrilateral is cyclic. But you can also use this property in reverse: if you know a quadrilateral is cyclic, you can conclude that opposite angles sum to $180°$.

4. Incorrect vector notation in geometry proofs. When writing vector proofs, always use position vectors (e.g. $\vec{OA}$, $\vec{OB}$) or clearly define your notation. Mixing free vectors and position vectors leads to sign errors.

5. Not rationalising the denominator in coordinate geometry answers. In DSE, answers involving surds should have rationalised denominators. For example, write $\dfrac{1}{\sqrt{2}}$ as $\dfrac{\sqrt{2}}{2}$.

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DSE Exam-Style Questions

DSE-1

$ABCD$ is a cyclic quadrilateral with $AB = AC$ and $AD$ produced to $E$ such that $CE$ is a tangent to the circle at $C$.

(a) Prove that $\angle ABC = \angle ACE$. (3 marks) (b) If $\angle BAC = 50°$ and $\angle ABC = 65°$, find $\angle ADC$. (3 marks)

Solution:

(a) Since $ABCD$ is cyclic: $\angle ABC + \angle ADC = 180°$.

The angle between tangent $CE$ and chord $AC$ equals the angle in the alternate segment:

$\angle ACE = \angle ABC$. (This is the alternate segment theorem.)

(b) $\angle ABC = 65°$ (given).

In triangle $ABC$: $\angle BCA = 180° - 50° - 65° = 65°$.

Since $ABCD$ is cyclic: $\angle ADC = 180° - \angle ABC = 180° - 65° = 115°$.

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DSE-2

The vertices of triangle $ABC$ are $A(1, 2)$, $B(5, 4)$, and $C(3, 8)$.

(a) Find the equation of the perpendicular bisector of $AB$. (4 marks) (b) The perpendicular bisector of $AB$ meets the perpendicular bisector of $AC$ at point $O$. Find the coordinates of $O$, the circumcentre of triangle $ABC$. (4 marks) (c) Find the radius of the circumcircle. (2 marks)

Solution:

(a) Midpoint of $AB$: $M = \left(\dfrac{1+5}{2},\; \dfrac{2+4}{2}\right) = (3, 3)$.

Slope of $AB$: $m_{AB} = \dfrac{4 - 2}{5 - 1} = \dfrac{1}{2}$.

Slope of perpendicular bisector: $m = -2$.

Equation: $y - 3 = -2(x - 3) \implies y = -2x + 9$.

(b) Midpoint of $AC$: $N = \left(\dfrac{1+3}{2},\; \dfrac{2+8}{2}\right) = (2, 5)$.

Slope of $AC$: $m_{AC} = \dfrac{8 - 2}{3 - 1} = 3$.

Slope of perpendicular bisector of $AC$: $m = -\dfrac{1}{3}$.

Equation: $y - 5 = -\dfrac{1}{3}(x - 2) \implies y = -\dfrac{1}{3}x + \dfrac{17}{3}$.

Intersection: $-2x + 9 = -\dfrac{1}{3}x + \dfrac{17}{3}$.

$-6x + 27 = -x + 17 \implies -5x = -10 \implies x = 2$.

$y = -2(2) + 9 = 5$.

Circumcentre: $O = (2, 5)$.

(c) Radius $= OA = \sqrt{(2-1)^2 + (5-2)^2} = \sqrt{1 + 9} = \sqrt{10}$.

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DSE-3

In triangle $PQR$, $PQ = 7$ cm, $PR = 5$ cm, and $\angle QPR = 60°$.

(a) Find $QR$. (3 marks) (b) Find the area of triangle $PQR$. (2 marks) (c) Find the length of the perpendicular from $P$ to $QR$. (2 marks)

Solution:

(a) By the cosine rule:

$QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos 60° = 49 + 25 - 2 \times 7 \times 5 \times \frac{1}{2} = 74 - 35 = 39$

$QR = \sqrt{39} \text{ cm}$

(b) Area $= \dfrac{1}{2} \times PQ \times PR \times \sin 60° = \dfrac{1}{2} \times 7 \times 5 \times \dfrac{\sqrt{3}}{2} = \dfrac{35\sqrt{3}}{4}$ cm$^2$.

(c) Area $= \dfrac{1}{2} \times QR \times h$ where $h$ is the perpendicular from $P$ to $QR$.

$\frac{35\sqrt{3}}{4} = \frac{1}{2} \times \sqrt{39} \times h$

$h = \frac{35\sqrt{3}}{2\sqrt{39}} = \frac{35\sqrt{3}}{2\sqrt{39}} \cdot \frac{\sqrt{39}}{\sqrt{39}} = \frac{35\sqrt{117}}{78} = \frac{35 \times 3\sqrt{13}}{78} = \frac{35\sqrt{13}}{26} \text{ cm}$

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DSE-4

The position vectors of points $A$, $B$, $C$ are $\mathbf{'\{'}a{'\}'}$, $\mathbf{'\{'}b{'\}'}$, $\mathbf{'\{'}c{'\}'}$ respectively. Point $D$ is such that $\vec{AD} = \dfrac{1}{3}\vec{AC}$.

(a) Express $\vec{OD}$ in terms of $\mathbf{'\{'}a{'\}'}$ and $\mathbf{'\{'}c{'\}'}$. (1 mark) (b) If $E$ is the midpoint of $BC$, prove that $A$, $D$, $E$ are collinear. (4 marks)

Solution:

(a) $\vec{AD} = \dfrac{1}{3}\vec{AC} = \dfrac{1}{3}(\mathbf{'\{'}c{'\}'} - \mathbf{'\{'}a{'\}'})$.

$\vec{OD} = \vec{OA} + \vec{AD} = \mathbf{'\{'}a{'\}'} + \frac{1}{3}(\mathbf{'\{'}c{'\}'} - \mathbf{'\{'}a{'\}'}) = \frac{2}{3}\mathbf{'\{'}a{'\}'} + \frac{1}{3}\mathbf{'\{'}c{'\}'}$

(b) $\vec{OE} = \dfrac{1}{2}(\mathbf{'\{'}b{'\}'} + \mathbf{'\{'}c{'\}'})$.

$\vec{AE} = \vec{OE} - \vec{OA} = \frac{1}{2}(\mathbf{'\{'}b{'\}'} + \mathbf{'\{'}c{'\}'}) - \mathbf{'\{'}a{'\}'}$

$\vec{AD} = \frac{1}{3}(\mathbf{'\{'}c{'\}'} - \mathbf{'\{'}a{'\}'})$

For collinearity, $\vec{AE}$ must be a scalar multiple of $\vec{AD}$. This requires more information about the relationship between $\mathbf{'\{'}a{'\}'}$, $\mathbf{'\{'}b{'\}'}$, and $\mathbf{'\{'}c{'\}'}$. If $D$ divides $AE$ in some ratio, we can show:

$\vec{AD} = \dfrac{1}{3}\vec{AC}$ means $D$ divides $AC$ in ratio $1:2$.

$E$ is the midpoint of $BC$.

By the converse of the midpoint theorem or using mass points: assign mass $2$ at $A$ and mass $1$ at $C$, giving the centre of mass at $D$ on $AC$. Similarly, mass $1$ at $B$ and $1$ at $C$ gives $E$ on $BC$.

Consider $\vec{DE} = \vec{OE} - \vec{OD} = \dfrac{1}{2}(\mathbf{'\{'}b{'\}'} + \mathbf{'\{'}c{'\}'}) - \dfrac{2}{3}\mathbf{'\{'}a{'\}'} - \dfrac{1}{3}\mathbf{'\{'}c{'\}'} = \dfrac{1}{2}\mathbf{'\{'}b{'\}'} + \dfrac{1}{6}\mathbf{'\{'}c{'\}'} - \dfrac{2}{3}\mathbf{'\{'}a{'\}'}$.

This shows collinearity only if additional conditions are given. The question likely assumes $D$ lies on the median from $A$, which it does since $D$ is on $AC$ and $E$ is on $BC$.

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DSE-5

A circle $C$ has equation $x^2 + y^2 - 8x + 6y + 9 = 0$.

(a) Find the centre and radius of $C$. (3 marks) (b) Find the equation of the tangent to $C$ at the point $(5, -2)$. (4 marks) (c) The tangent in part (b) meets the $x$-axis at $T$. Find the coordinates of $T$. (2 marks)

Solution:

(a) $(x^2 - 8x + 16) + (y^2 + 6y + 9) = -9 + 16 + 9$.

$(x - 4)^2 + (y + 3)^2 = 16$.

Centre: $(4, -3)$, Radius: $4$.

(b) The tangent at $(5, -2)$ is perpendicular to the radius joining $(4, -3)$ and $(5, -2)$.

Slope of radius: $m_r = \dfrac{-2 - (-3)}{5 - 4} = 1$.

Slope of tangent: $m_t = -1$.

Equation: $y - (-2) = -1(x - 5) \implies y + 2 = -x + 5 \implies x + y - 3 = 0$.

(c) On the $x$-axis, $y = 0$: $x - 3 = 0 \implies x = 3$.

$T = (3, 0)$.