Geometry, Trig & Calculus

Analytical and Euclidean Geometry: Circles and Lines

Grade 10 · Grade 11 · Grade 12

  • ✓By the end of this lesson students will be able to calculate the distance between two points, the midpoint of a line segment, and the gradient of a line.
  • ✓By the end of this lesson students will be able to determine the equation of a straight line and apply conditions for parallel and perpendicular lines.
  • ✓By the end of this lesson students will be able to apply the properties of circles, including the equation of a circle, to solve analytical geometry problems.
  • ✓By the end of this lesson students will be able to prove and apply the theorems related to angles, chords, and tangents in circles.
  • ✓By the end of this lesson students will be able to solve problems involving the inclination of a line and the properties of cyclic quadrilaterals.

Key concepts

Distance Formula

The distance between two points A(x₁, y₁) and B(x₂, y₂) in the Cartesian plane.

d = √((x₂ - x₁)² + (y₂ - y₁)²)
Midpoint Formula

The coordinates of the midpoint M of a line segment joining A(x₁, y₁) and B(x₂, y₂).

M = ((x₁ + x₂)/2 ; (y₁ + y₂)/2)
Gradient of a Line

The measure of the steepness of a line, denoted by 'm'. It represents the change in y divided by the change in x.

m = (y₂ - y₁)/(x₂ - x₁)
Equation of a Straight Line

The algebraic representation of a straight line. Common forms include gradient-intercept form (y = mx + c) and point-gradient form (y - y₁ = m(x - x₁)).

y = mx + c OR y - y₁ = m(x - x₁)
Parallel and Perpendicular Lines

Two lines are parallel if their gradients are equal (m₁ = m₂). Two lines are perpendicular if the product of their gradients is -1 (m₁ × m₂ = -1).

m₁ = m₂ (for parallel); m₁ × m₂ = -1 (for perpendicular)
Inclination of a Line

The angle (θ) that a line makes with the positive x-axis, measured anti-clockwise. The gradient 'm' is related to the inclination by tan θ = m.

tan θ = m
Equation of a Circle

The algebraic representation of a circle with centre (a, b) and radius r. If the centre is at the origin (0,0), the equation simplifies to x² + y² = r².

(x - a)² + (y - b)² = r²
Circle Theorem 1: Perpendicular from Centre to Chord

The line drawn from the centre of a circle perpendicular to a chord bisects the chord. Conversely, the line from the centre to the midpoint of a chord is perpendicular to the chord.

Circle Theorem 2: Angle at Centre and Circumference

The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circumference.

Circle Theorem 3: Angles in the Same Segment

Angles subtended by the same arc in the same segment of a circle are equal.

Circle Theorem 4: Angle in a Semi-Circle

The angle subtended by a diameter at any point on the circumference is a right angle (90°).

Circle Theorem 5: Cyclic Quadrilateral Properties

The opposite angles of a cyclic quadrilateral are supplementary (add up to 180°). The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Circle Theorem 6: Tangent Perpendicular to Radius

The tangent to a circle is perpendicular to the radius (or diameter) at the point of contact.

Circle Theorem 7: Tangents from a Common Point

Two tangents drawn from an external point to a circle are equal in length.

Circle Theorem 8: Tan-Chord Theorem

The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle in the alternate segment.

Key facts to remember

  • 1Distance Formula: d = √((x₂ - x₁)² + (y₂ - y₁)²)
  • 2Midpoint Formula: M = ((x₁ + x₂)/2 ; (y₁ + y₂)/2)
  • 3Gradient of a Line: m = (y₂ - y₁)/(x₂ - x₁)
  • 4Equation of a Straight Line: y = mx + c or y - y₁ = m(x - x₁)
  • 5Parallel lines have equal gradients (m₁ = m₂). Perpendicular lines have gradients whose product is -1 (m₁ × m₂ = -1).
  • 6Inclination of a line: tan θ = m, where θ is the angle with the positive x-axis.
  • 7Equation of a Circle with centre (a, b) and radius r: (x - a)² + (y - b)² = r².
  • 8Key Circle Theorems: Angle at centre = 2 × angle at circumference; Angles in same segment are equal; Angle in semi-circle = 90°; Opposite angles of cyclic quad are supplementary; Tangent ⊥ Radius; Tan-chord theorem.

Worked examples

Example 1

Consider the points A(-2; 3), B(4; 5) and C(2; -1). Calculate the length of AB, the gradient of BC, and the equation of the line perpendicular to BC passing through A.

I1. Calculate the length of AB using the distance formula:
IId_AB = √((x_B - x_A)² + (y_B - y_A)²)
IIId_AB = √((4 - (-2))² + (5 - 3)²)
IVd_AB = √((6)² + (2)²)
Vd_AB = √(36 + 4)
VId_AB = √40
VIId_AB = 2√10 units
VIII2. Calculate the gradient of BC:
9m_BC = (y_C - y_B)/(x_C - x_B)
10m_BC = (-1 - 5)/(2 - 4)
11m_BC = (-6)/(-2)
12m_BC = 3
133. Determine the gradient of the line perpendicular to BC:
14For perpendicular lines, m_perpendicular × m_BC = -1
15m_perpendicular × 3 = -1
16m_perpendicular = -1/3
174. Find the equation of the line perpendicular to BC passing through A(-2; 3) using y - y₁ = m(x - x₁):
18y - 3 = (-1/3)(x - (-2))
19y - 3 = (-1/3)(x + 2)
203(y - 3) = -1(x + 2)
213y - 9 = -x - 2
22x + 3y - 7 = 0

Answer

Length of AB = 2√10 units. Gradient of BC = 3. Equation of the perpendicular line is x + 3y - 7 = 0.

Remember to state the formula before substitution and simplify surds where possible.

Example 2

A circle has its centre at M(1; -2) and passes through the point P(4; 2). Determine the equation of the circle and the equation of the tangent to the circle at point P.

I1. Determine the radius (r) of the circle. The radius is the distance between the centre M(1; -2) and the point P(4; 2).
IIr² = (x_P - x_M)² + (y_P - y_M)²
IIIr² = (4 - 1)² + (2 - (-2))²
IVr² = (3)² + (4)²
Vr² = 9 + 16
VIr² = 25
VIIr = 5 units
VIII2. Write the equation of the circle using the centre M(1; -2) and r² = 25:
9(x - a)² + (y - b)² = r²
10(x - 1)² + (y - (-2))² = 25
11(x - 1)² + (y + 2)² = 25
123. Determine the gradient of the radius MP:
13m_MP = (y_P - y_M)/(x_P - x_M)
14m_MP = (2 - (-2))/(4 - 1)
15m_MP = (4)/(3)
164. Determine the gradient of the tangent at P. The tangent is perpendicular to the radius at the point of contact.
17m_tangent × m_MP = -1
18m_tangent × (4/3) = -1
19m_tangent = -3/4
205. Find the equation of the tangent using the point P(4; 2) and m_tangent = -3/4:
21y - y₁ = m(x - x₁)
22y - 2 = (-3/4)(x - 4)
234(y - 2) = -3(x - 4)
244y - 8 = -3x + 12
253x + 4y - 20 = 0

Answer

Equation of the circle: (x - 1)² + (y + 2)² = 25. Equation of the tangent at P: 3x + 4y - 20 = 0.

The radius is perpendicular to the tangent at the point of contact. This is a crucial theorem for tangent problems.

Example 3

In the diagram, O is the centre of the circle. A, B, C and D are points on the circumference. Chords AC and BD intersect at E. AB is parallel to DC. If ∠BOC = 100° and ∠OAC = 30°, calculate the size of ∠ADC and ∠AEC. Provide reasons for your statements.

I1. Calculate ∠ADC:
IIIn ΔAOC, OA = OC (radii)
IIITherefore, ∠OCA = ∠OAC = 30° (angles opposite equal sides)
IV∠AOC = 180° - (30° + 30°) = 120° (sum of angles in Δ)
V∠ADC = (1/2)∠AOC (angle at centre = 2 × angle at circumference)
VI∠ADC = (1/2) × 120° = 60°
VII2. Calculate ∠AEC:
VIIIFirst, find ∠BAC:
9∠BAC = ∠BDC (angles in same segment)
10∠BDC = (1/2)∠BOC (angle at centre = 2 × angle at circumference)
11∠BDC = (1/2) × 100° = 50°
12Therefore, ∠BAC = 50°
13Next, find ∠ACD:
14Since AB || DC, ∠ACD = ∠BAC (alternate angles, AB || DC)
15Therefore, ∠ACD = 50°
16Now, in ΔAEC:
17∠AEC = 180° - (∠EAC + ∠ECA) (sum of angles in Δ)
18∠AEC = 180° - (∠BAC + ∠ACD)
19∠AEC = 180° - (50° + 50°)
20∠AEC = 180° - 100° = 80°

Answer

∠ADC = 60°. ∠AEC = 80°.

Always provide a reason for each step in Euclidean geometry. Break down complex problems into smaller, manageable parts.

Common mistakes

  • ✗Swapping x and y coordinates in formulas, especially in the gradient or midpoint formula.
  • ✗Incorrectly applying the negative reciprocal for perpendicular gradients (e.g., using 1/m instead of -1/m).
  • ✗Forgetting to square the radius in the equation of a circle, or incorrectly taking the square root of r².
  • ✗Not providing reasons for statements in Euclidean geometry proofs, which results in loss of marks.
  • ✗Confusing the angle at the centre with the angle at the circumference, or applying the theorem incorrectly (e.g., using the wrong arc).
  • ✗Misidentifying the alternate segment for the Tan-Chord Theorem.

Exam tips

  • ★Always draw a clear diagram for analytical geometry problems, labelling all given points and information.
  • ★For Euclidean geometry, draw the diagram accurately and mark all given information. Redraw if the diagram is too cluttered.
  • ★Write down the formula you are using before substituting values in analytical geometry. This helps avoid errors and earns method marks.
  • ★For Euclidean geometry proofs, state the theorem (reason) for every geometric statement you make. Use standard abbreviations (e.g., 'sum of angles in Δ', 'angles opp equal sides').
  • ★Check your calculations carefully, especially when dealing with negative numbers or fractions.

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