A Level Algebra and Functions, A Level Maths

A Level Circles – Maths Revision

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:

x^2 + y^2 = r^2

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 𝑥2 with (𝑥 – 𝑎)2
  • To move the centre 𝑏 up, we replace 𝑦2 with (𝑦 – 𝑏)2

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

(x - a)^2 + (y-b)^2=r^2

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 𝑥2 + 4𝑥 + 𝑦2 – 8𝑦 + 11 = 0.
a. Find the coordinates of the centre and the radius of the circle.

a. The 𝑥 terms are 𝑥2 + 4𝑥 so begin by writing this expression in completed square form:

x^2 + 4x = (x+2)^2 - 4

The 𝑦 terms are 𝑦2 – 8𝑦. In completed square form:

y^2-8y=(y-4)^2-16

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

\begin{aligned} &(x+2)^2-4+(y-4)^2-16+11=0 \\ &(x+2)^2+(y-4)^2-9=0\\&(x+2)^2+(y-4)^2=9 \end{aligned}

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.

\begin{aligned} &(x + 2)^2 + (y-4)^2=9\\&(1+2)^2 +(4-4)^2=9\\&3^2+0^2=9\end{aligned}

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 (𝑥1, 𝑦1) and (𝑥2, 𝑦2) is given by:

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

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 = 𝑟2 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:

\begin{aligned} r &= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ &=\sqrt{(0-3)^2+(1-0)^2}\\&=\sqrt{10}\end{aligned} .

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

\begin{aligned} &(x-0)^2+(y-1)^2=(\sqrt{10})^2\\&x^2+(y-1)^2=10\end{aligned}

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:

x^2 +(y-1)^2=r^2

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

\begin{aligned} &3^2 +(0-1)^2=r^2\\&10=r^2 \end{aligned}

Again, the equation is 𝑥2 + (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:

y = x + 3

Then, substitute this into the equation of the circle.

\begin{aligned}&(x+1)^2+(x+3-5)^2=8\\&(x+1)^2+(x-2)^2=8\\&x^2+2x+1+x^2-4x+4=8\\&2x^2-2x-3=0\end{aligned}

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

x = \frac{1\,+\,\sqrt{7}}{2}, x = \frac{1\,-\,\sqrt{7}}{2}

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

\left(\frac{1 \,+\, \sqrt{7}}{2},\frac{7\,+\,\sqrt{7}}{2}\right),\left(\frac{1\,-\,\sqrt{7}}{2},\frac{7\,-\,\sqrt{7}}{2}\right)

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), 𝑚𝑟. You can then use this to work out the gradient of the tangent, 𝑚𝑡.

\begin{aligned} m_r &=\frac{5\,-\,2}{1\,-\,3}\\&=-\frac{3}{2}\end{aligned}

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.

\begin{aligned} &m_t &=\frac{2}{3}\end{aligned}

By substituting 𝑚𝑡 and (1, 5) into the formula (𝑦 – 𝑦1) = 𝑚(𝑥 – 𝑥1), or by some other means, find the equation of the tangent:

\begin{aligned} &(y-5)=\frac{2}{3}(x-1)\\&3y-15=2x-2\\&3y-2x-13=0\end{aligned}

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. 𝑥2 + 6𝑥 + 𝑦2 + 2𝑦 + 5 = 0
b. 𝑥2 + 𝑦2 – 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

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

a. (𝑥 + 3)2 + (𝑦 – 4)2 = 16

\boldsymbol{\begin{aligned} &(-3, 4), r=4 \end{aligned}}

b. (𝑥 – 2)2 + (𝑦 + 5)2 = 36

\boldsymbol{\begin{aligned} (2,-5), r =6 \end{aligned}}

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. 𝑥2 + 6𝑥 + 𝑦2 + 2𝑦 + 5 = 0

\boldsymbol{\begin{aligned} &(x+3)^2-9+(y+1)^2-1+5=0\\&(x+3)^2+(y+1)^2-5=0\\&(x+3)^2+(y+1)^2=5\\&(-3,-1),r=\sqrt{5}\end{aligned}}

b. 𝑥2 + 𝑦2 – 8𝑥 + 5𝑦 – 1 = 0

\boldsymbol{\begin{aligned} &(x-4)^2-16+(y+\frac{5}{2})^2-\frac{25}{4}-1=0\\&(x-4)^2+(y+\frac{5}{2})^2=\frac{93}{4}\\&(4, -\frac{5}{2}), r = \frac{\sqrt{93}}{2}\end{aligned}}

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

The equation of the circle is:

\boldsymbol{(x+1)^2+(y-2)^2=r^2}

Substitute 𝑥 = 5 and 𝑦 = 8 into this equation:

\boldsymbol{\begin{aligned}&(5+1)^2+(8-2)^2=r^2\\&72=r^2 \end{aligned}}

The equation is:

\boldsymbol{(x+1)^2+(y-2)^2=72}

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.

Substitute 𝑦 = 1 – 𝑥 into the equation of the circle and solve for 𝑥:

\boldsymbol{\begin{aligned}&(x-1)^2+(1-x+4)^2=8\\&2x^2-12x+18=0\\&(x-3)^2=0\\&x=3\\&y=1-3=-2\\&(3, -2)\end{aligned}}

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.

The centre of the circle is (-3, 1). The line joining this point to (5, 2) is the radius.

The gradient of the radius is:

\boldsymbol{m=\frac{2\,-\,1}{5\,-\,(-3)}=\frac{1}{8}}

Since the tangent and radius are perpendicular, the gradient of the tangent is -8.

This gives an equation for the circle of:

\boldsymbol{\begin{aligned}&y-y_1=m(x-x_1)\\&y-2=-8(x-5)\\&y-2=-8x+40\\&y+8x-42=0\end{aligned}}

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