Straight Lines and Circles — CSEC Add Maths notes

Points and distances

Both of these come straight from a right-angled triangle drawn between the two points.

The midpoint is just the average of the coordinates. The distance is Pythagoras.

Gradients, parallel and perpendicular

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Negative reciprocal means two changes, not one: flip the fraction and change the sign. The perpendicular to a line of gradient 3 has gradient $-\tfrac13$ , not 3, not $-3$ , and not $\tfrac13$ .

The equation of a straight line

Given a gradient and any point on the line:

$y - y_1 = m(x - x_1)$

Given two points, find $m$ first, then use either point — both give the same line.

The equation of a circle

Read the signs carefully: centre $(2, -3)$ gives $(x-2)^2 + (y+3)^2$ , because subtracting $-3$ is adding 3. And the right-hand side is $r^2$ , not $r$ .

To build the equation you need the centre and the radius. If you are given the centre and a point on the circle, the radius is the distance between them — and since you only ever need $r^2$ , you can stop before taking the square root.

The general form

Expanding the equation above and collecting terms gives the other form you will meet:

$x^2 + y^2 + 2gx + 2fy + c = 0$

The coefficients of $x$ and $y$ are $2g$ and $2f$ , so halve them and change the sign to get the centre. For $x^2 + y^2 - 6x + 4y - 12 = 0$ : half of $-6$ is $-3$ , so $g = -3$ and the $x$ -coordinate of the centre is $+3$ . Similarly $f = 2$ gives $-2$ . The centre is $(3, -2)$ and the radius is $\sqrt{9 + 4 + 12} = 5$ .

You can also complete the square in $x$ and in $y$ to convert one form into the other, which is worth practising because it explains where the formula comes from.

Tangents to a circle

The key geometric fact:

So to find the tangent at a point $P$ on a circle with centre $C$ :

Find the gradient of $CP$ .
Take its negative reciprocal — that is the gradient of the tangent.
Use $y - y_1 = m(x - x_1)$ with the point $P$ .

Does a line meet a circle?

Substitute the line into the circle to eliminate one variable. You are left with a quadratic, and its discriminant answers the question:

$\Delta > 0$ — the line cuts the circle at two points;
$\Delta = 0$ — the line is a tangent;
$\Delta < 0$ — the line misses the circle entirely.

This is the same discriminant test used for a line and a parabola. "Show that the line is a tangent to the circle" always means "show the discriminant is zero", and the repeated root it gives you is the $x$ -coordinate of the point of contact.