13.4 The equation of a circle and tangents — Learn

Prerequisites: You can use coordinates, Pythagoras' theorem and gradients of straight lines.

Objectives: You will use circle equations, test points and find tangent equations using perpendicular gradients.

Circle equations

A circle with centre $(0,0)$ and radius $r$ has equation

The two squared distances from the centre add to the squared radius.

Example 1. State the centre and radius of $x^2+y^2=25$.

Compare with the equation for a circle centred at the origin:

Take the positive square root because a radius is a length:

Example 2. Check whether $(3,4)$ lies on $x^2+y^2=25$.

Substitute both coordinates:

The equation is true, so the point lies on the circle.

Tangents

A tangent touches a circle at one point. At that point, the tangent is perpendicular to the radius. For non-vertical lines, perpendicular gradients multiply to $-1$.

Example 3. Find the tangent equation at $(3,4)$ to the circle $x^2+y^2=25$.

The radius gradient from $(0,0)$ to $(3,4)$ is $4/3$:

Use the negative reciprocal for the tangent gradient:

Use $y-y_1=m(x-x_1)$ through $(3,4)$:

Watch out. The equation in this section is for a circle centred at the origin. Check that both squared terms are $x^2$ and $y^2$ before reading off the radius.

Progress check

A circle has centre $(1,2)$ and a radius with gradient $2$. Predict the tangent gradient and check that their product is $-1$.

You should now be able to

• find the centre and radius from a circle equation;
• write the equation of a circle centred at the origin from its radius;
• test whether a point lies on a circle;
• find a tangent gradient and equation using perpendicular lines.