Straight lines, gradients, parallel and perpendicular conditions, the equation of a circle, the relationship between a tangent and the radius, and parametric equations of curves.

What this dot point is asking

OCR wants you to find the equation of a straight line from points or a gradient, use the conditions for parallel and perpendicular lines, find the midpoint and distance between two points, work with the equation of a circle (including completing the square to find centre and radius), use the geometric facts that a tangent meets the radius at a right angle and that the perpendicular from the centre bisects a chord, and convert between parametric and Cartesian equations.

The answer

Straight lines

The equation of a circle

A circle with centre $(a, b)$ and radius $r$ has equation $(x - a)^2 + (y - b)^2 = r^2$. When a circle is given in expanded form, complete the square in $x$ and in $y$ to recover the centre and radius.

Tangent and chord geometry

Two facts unlock most circle problems:

• A tangent is perpendicular to the radius at the point of contact, so the tangent gradient is the negative reciprocal of the radius gradient.
• The perpendicular from the centre bisects a chord, so the centre lies on the perpendicular bisector of any chord.

Parametric equations

A curve can be given as $x = f(t)$, $y = g(t)$. To find a Cartesian equation, eliminate $t$: solve one equation for $t$ and substitute, or use an identity. Always carry any domain restriction on $t$ through to the Cartesian curve.

Examples in context

A perpendicular bisector

Finding where a line meets a circle

To find the intersection of a line and a circle, substitute the line into the circle equation and solve the resulting quadratic. The discriminant of that quadratic tells you whether the line is a tangent (one solution, $\Delta = 0$), a secant cutting the circle twice ($\Delta > 0$), or misses it entirely ($\Delta < 0$). This is a favourite OCR synoptic link between coordinate geometry and the discriminant.

Using a circle property to find a centre

If you know three points on a circle, the centre is equidistant from all of them, so it lies on the perpendicular bisectors of the chords joining them. Finding two perpendicular bisectors and solving them simultaneously locates the centre, after which the radius is the distance to any of the three points. This blends the line tools (midpoint, perpendicular gradient) with the circle definition.

Try this

Q1. Find the equation of the line through $(2, -1)$ perpendicular to $y = 2x + 3$. [3 marks]

• Cue. Gradient $-\tfrac{1}{2}$, so $y + 1 = -\tfrac{1}{2}(x - 2)$.

Q2. State the centre and radius of $(x - 4)^2 + (y + 1)^2 = 16$. [2 marks]

• Cue. Centre $(4, -1)$, radius $4$.