Are you looking for info on circles (Maths revision for A level)? A circle, with a centre at the origin and a radius of 𝑟, can be described with the equation:

You can generalise this to find the equation of a circle with a centre at any point by applying a transformation to the above graph.

- To move the centre 𝑎 to the right, we replace 𝑥
^{2}with (𝑥 – 𝑎)^{2} - To move the centre 𝑏 up, we replace 𝑦
^{2}with (𝑦 – 𝑏)^{2}

This gives the equation for a circle with a centre at (𝑎, 𝑏) and a radius of 𝑟:

If a question does not give the equation of a circle in this form, you will need to complete the square for 𝑥 and 𝑦 to convert it to the general form.

Example Question 1

**A circle has equation 𝑥 ^{2} + 4𝑥 + 𝑦^{2} – 8𝑦 + 11 = 0.**

a. Find the coordinates of the centre and the radius of the circle.

a. The 𝑥 terms are 𝑥^{2} + 4𝑥 so begin by writing this expression in completed square form:

The 𝑦 terms are 𝑦^{2} – 8𝑦. In completed square form:

Put this all back into the original equation, collect the constants and rearrange into the equation for a circle:

The centre is (-2, 4). The radius squared is 9, so the radius is 3 units.

**b. Show that the point with coordinates (1, 4) lies on the circumference of this circle.**

Substitute 𝑥 = 1 and 𝑦 = 4 into the equation of the circle found in part **a**.

**c. Does the circle intersect the axes? Explain your reasoning.**

The centre is (-2, 4) and its radius is 3 units. Since the radius, 3, is greater than the smallest distance of the centre from the origin, 2, the circle must intersect the 𝑦-axis.

It is important to realise that, given the coordinates of the centre of the circle and a point on its circumference, you can find the length of the radius by using the distance formula.

This formula, which is a direct result of Pythagoras’ theorem, tells us that the distance, 𝑑, between two points (𝑥_{1}, 𝑦_{1}) and (𝑥_{2}, 𝑦_{2}) is given by:

Example Question 2

**The point with coordinates (3, 0) lies on a circle with centre (0, 1). Find the equation of this circle, giving your answer in the form (𝑥 – 𝑎) ^{2} + (𝑦 – 𝑏)^{2} = 𝑟^{2} where 𝑎, 𝑏 and 𝑟 are constants to be found.**

Method 1

The radius is the line segment joining (3, 0) and (0, 1). Use the distance formula to find its length:

.

As you are given the centre coordinates, the equation is:

Therefore, 𝑎 = 0, 𝑏 = 1 and 𝑟 = √10

Method 2

Since the centre is (0, 1), the equation of the circle can be provisionally written as:

Since the circumference of the circle passes through (3, 0), you can substitute 𝑥 = 3 and 𝑦 = 0 into this equation:

Again, the equation is 𝑥^{2} + (𝑦 – 1)^{2} = 10

You can find the equation of the points where a line or curve meets the circumference of a circle by solving their respective equations simultaneously.

Example Question 3

**Find the exact coordinates of the points of intersection of the line with equation 𝑦 – 𝑥 = 3 and the circle whose equation is (𝑥 + 1) ^{2} + (𝑦 – 5)^{2} = 8.**

To solve the equations simultaneously, begin by rearranging 𝑦 – 𝑥 = 3 to make 𝑦 the subject:

Then, substitute this into the equation of the circle.

Solving this (remember, you can use your calculator) gives you:

Substitute these values back into the equation 𝑦 – 𝑥 = 3 to get the corresponding 𝑦-values (be careful not to mix these up!):

Finally, you can use circle theorems to solve problems involving the equation of a circle.

The key ones to recall are:

- The tangent and radius of a circle meet at 90°
- The angle subtended by the diameter of a semicircle is 90°. (Note: this is sometimes referred to as “The angle in a semicircle is 90°.”)

Example Question 4

**A circle with centre (3, 2) has a tangent at the point with coordinates (1, 5). Work out the equation of this line, giving your answer in the form 𝑎𝑦 + 𝑏𝑥 + 𝑐 = 0.**

The tangent and radius of a circle meet at 90° so begin by working out the gradient of the radius between (3, 2) and (1, 5), 𝑚_{𝑟}. You can then use this to work out the gradient of the tangent, 𝑚_{𝑡}.

The product of the gradients of two perpendicular lines is -1 or, alternatively, you can say the gradient of the tangent is the negative reciprocal of the gradient of the radius.

By substituting 𝑚_{𝑡} and (1, 5) into the formula (𝑦 – 𝑦_{1}) = 𝑚(𝑥 – 𝑥_{1}), or by some other means, find the equation of the tangent:

Practice Questions

1. Write down the coordinates of the centre and the radius of each circle.

a. (𝑥 + 3)^{2} + (𝑦 – 4)^{2} = 16

b. (𝑥 – 2)^{2} + (𝑦 + 5)^{2} = 36

2. For each circle, work out the radius and the coordinates of the centre. Where necessary, give your answer as a surd in its simplest form.

a. 𝑥^{2} + 6𝑥 + 𝑦^{2} + 2𝑦 + 5 = 0

b. 𝑥^{2} + 𝑦^{2} – 8𝑥 + 5𝑦 – 1 = 0

3. The circumference of a circle with centre (-1, 2) passes through the point with coordinates (5, 8). Find the equation of this circle.

4. The circle whose equation is (𝑥 – 1)^{2} + (𝑦 + 4)^{2} = 8 has a tangent with equation 𝑦 + 𝑥 = 1. Work out the point at which this tangent meets the circumference of the circle.

5. A circle with equation (𝑥 + 3)^{2} + (𝑦 – 1)^{2} =√65 has a tangent at (5, 2). Work out the equation of the tangent, giving your answer in the form 𝑎𝑦 + 𝑏𝑥 + 𝑐 = 0.

Answers

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