Lines and Circles — Algebraic Methods & Geometric Properties (9709 P1) — Mathematics

Introduction

Coordinate geometry problems on CAIE 9709 Paper 1 frequently combine the equation of a circle with properties of straight lines. You are expected to move fluently between algebra and classical circle geometry — for example, finding where a line meets a circle, proving a line is tangent, or using the fact that a radius meets a tangent at a right angle. These questions carry significant marks and reward methodical, well-justified working. Implicit differentiation is not required; every gradient you need can be found from coordinates alone.

Core Concept

The Circle and Lines — Setting Up

Recall from the prerequisite topic that a circle with centre $C(a, b)$ and radius $r$ has equation:

(x - a)^2 + (y - b)^2 = r^2

A straight line can interact with a circle in three ways: no intersection (line misses the circle), tangency (one point of contact), or two distinct intersections. The algebraic test is the discriminant of the quadratic formed by substituting the line into the circle equation.

Discriminant $\Delta$	Geometric situation
$\Delta > 0$	Line intersects circle at two distinct points
$\Delta = 0$	Line is a tangent to the circle
$\Delta < 0$	Line does not meet the circle

Three Key Geometric Properties (Examinable)

1. Tangent perpendicular to radius. At any point $P$ on the circle, the tangent at $P$ is perpendicular to the radius $CP$ . This means:

m_{\text{radius}} \times m_{\text{tangent}} = -1

2. Angle in a semicircle. If $AB$ is a diameter of a circle, then for any point $P$ on the circle, $\angle APB = 90°$ . Algebraically: $\overrightarrow{PA} \cdot \overrightarrow{PB} = 0$ .

3. Symmetry — perpendicular from centre to a chord. The perpendicular from the centre of a circle to any chord bisects the chord. Equivalently, the line joining the centre to the midpoint of a chord is perpendicular to that chord.

Finding Intersections of a Line and a Circle

Step 1. Express $y$ (or $x$ ) from the line equation.
Step 2. Substitute into the circle equation to get a quadratic.
Step 3. Solve the quadratic; check the discriminant if needed.
Step 4. Back-substitute to find the full coordinates.

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