7.1 Coordinate geometry, equation of tangent, transformations

Co-ordinate geometry

Equation of a circle

We know that every point on the circumference of a circle is the same distance away from the centre of the circle. Consider a point $(x_1,y_1)$ on the circumference of a circle of radius $r$ with centre at $(x_0,y_0)$ .

In [link] , $▵OPQ$ is a right-angled triangle. Therefore, from the Theorem of Pythagoras, we know that:

But,

But, this same relation holds for any point $P$ on the circumference. In fact, the relation holds for all points $P$ on the circumference. Therefore, we can write:

for a circle with centre at $(x_0,y_0)$ and radius $r$ .

For example, the equation of a circle with centre $(0,0)$ and radius 4 is:

Equation of a tangent to a circle at a point on the circle

We are given that a tangent to a circle is drawn through a point $P$ with co-ordinates $(x_1,y_1)$ . In this section, we find out how to determine the equation of that tangent.

We start by making a list of what we know:

1. We know that the equation of the circle with centre $(x_0,y_0)$ and radius $r$ is ${(x-x_0)}^2+{(y-y_0)}^2=r^2$ .
2. We know that a tangent is perpendicular to the radius, drawn at the point of contact with the circle.

As we have seen in earlier grades, there are two steps to determining the equation of a straight line:

1. Calculate the gradient of the line, $m$ .
2. Calculate the $y$ -intercept of the line, $c$ .

The same method is used to determine the equation of the tangent. First we need to find the gradient of the tangent. We do this by finding the gradient of the line that passes through the centre of the circle and point $P$ (line $f$ in [link] ), because this line is a radius line and the tangent is perpendicular to it.