Equations of straight lines, parallel and perpendicular gradients, the equation of a circle, tangents and chords, and intersections of lines and curves.

What this dot point is asking

WJEC wants you to work with straight lines (gradient, intercepts, the various forms of the equation, parallel and perpendicular conditions) and circles (the standard equation, finding the centre and radius by completing the square, tangents and chords). You also find where lines and curves intersect by solving simultaneously. These results reappear in calculus, vectors and mechanics, so they must be automatic.

The answer

Straight lines

The gradient between two points is $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. The most useful forms of a line are:

The distance between two points is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, and the midpoint is $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.

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 given in expanded form $x^2 + y^2 + Dx + Ey + F = 0$, complete the square in $x$ and $y$ to recover the centre and radius.

Tangents, chords and intersections

A tangent touches the circle at exactly one point and is perpendicular to the radius drawn to that point. This radius-tangent property is the standard route to a tangent equation: find the gradient of the radius, take the negative reciprocal, then use the point of contact.

To find where a line meets a curve, solve their equations simultaneously. The number of intersection points matches the number of real solutions, which you can confirm with the discriminant when the result is a quadratic.

Examples in context

Example 1. Perpendicular bisector. Find the perpendicular bisector of the segment joining $A(2, 1)$ and $B(6, 9)$. The midpoint is $(4, 5)$ and the gradient of $AB$ is $\dfrac{9-1}{6-2} = 2$, so the perpendicular gradient is $-\dfrac{1}{2}$. The bisector is $y - 5 = -\dfrac{1}{2}(x - 4)$, which simplifies to $x + 2y - 14 = 0$. This is exactly the line of points equidistant from $A$ and $B$.

Example 2. Does a line cut a circle? Determine whether the line $y = x + 1$ meets the circle $x^2 + y^2 = 4$. Substituting gives $x^2 + (x+1)^2 = 4$, so $2x^2 + 2x - 3 = 0$. The discriminant is $4 + 24 = 28 > 0$, so the line cuts the circle at two points. The discriminant decides the geometry without solving fully.

Try this

Q1. Find the equation of the line through $(2, -1)$ perpendicular to $y = 3x + 4$. [3 marks]

• Cue. Perpendicular gradient is $-\tfrac{1}{3}$, so $y + 1 = -\tfrac{1}{3}(x - 2)$.

Q2. A circle has centre $(2, -3)$ and passes through $(5, 1)$. Find its equation. [3 marks]

• Cue. $r^2 = (5-2)^2 + (1+3)^2 = 9 + 16 = 25$, so $(x-2)^2 + (y+3)^2 = 25$.

Q3. Show that the line $y = 2x - 5$ is a tangent to the circle $x^2 + y^2 = 5$. [3 marks]

• Cue. Substitute to get $5x^2 - 20x + 20 = 0$; the discriminant is $400 - 400 = 0$, so there is one contact point.