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 π‘Ÿ, 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 π‘Ž to the right, we replace π‘₯2 with (π‘₯ – π‘Ž)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 (π‘Ž, 𝑏) and a radius of π‘Ÿ:

(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 π‘₯ and 𝑦 to convert it to the general form.


Example Question 1

A circle has equation π‘₯2 + 4π‘₯ + 𝑦2 – 8𝑦 + 11 = 0.
a. Find the coordinates of the centre and the radius of the circle.

a. The π‘₯ terms are π‘₯2 + 4π‘₯ 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 π‘₯ = 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 (π‘₯1, 𝑦1) and (π‘₯2, 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 (π‘₯ – π‘Ž)2 + (𝑦 – 𝑏)2 = π‘Ÿ2 where π‘Ž, 𝑏 and π‘Ÿ 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, π‘Ž = 0, 𝑏 = 1 and π‘Ÿ = √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 π‘₯ = 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 π‘₯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 – π‘₯ = 3 and the circle whose equation is (π‘₯ + 1)2 + (𝑦 – 5)2 = 8.

To solve the equations simultaneously, begin by rearranging 𝑦 – π‘₯ = 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 – π‘₯ = 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 90Β°
  • The angle subtended by the diameter of a semicircle is 90Β°. (Note: this is sometimes referred to as β€œThe angle in a semicircle is 90Β°.”)

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 π‘Žπ‘¦ + 𝑏π‘₯ + 𝑐 = 0.

The tangent and radius of a circle meet at 90Β° so begin by working out the gradient of the radius between (3, 2) and (1, 5), π‘šπ‘Ÿ. You can then use this to work out the gradient of the tangent, π‘šπ‘‘.

\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 π‘šπ‘‘ and (1, 5) into the formula (𝑦 – 𝑦1) = π‘š(π‘₯ – π‘₯1), 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. (π‘₯ + 3)2 + (𝑦 – 4)2 = 16
b. (π‘₯ – 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. π‘₯2 + 6π‘₯ + 𝑦2 + 2𝑦 + 5 = 0
b. π‘₯2 + 𝑦2 – 8π‘₯ + 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 (π‘₯ – 1)2 + (𝑦 + 4)2 = 8 has a tangent with equation 𝑦 + π‘₯ = 1. Work out the point at which this tangent meets the circumference of the circle.

5. A circle with equation (π‘₯ + 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 π‘Žπ‘¦ + 𝑏π‘₯ + 𝑐 = 0.


Answers

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

a. (π‘₯ + 3)2 + (𝑦 – 4)2 = 16

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

b. (π‘₯ – 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. π‘₯2 + 6π‘₯ + 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. π‘₯2 + 𝑦2 – 8π‘₯ + 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 π‘₯ = 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 (π‘₯ – 1)2 + (𝑦 + 4)2 = 8 has a tangent with equation 𝑦 + π‘₯ = 1. Work out the point at which this tangent meets the circumference of the circle.

Substitute 𝑦 = 1 – π‘₯ into the equation of the circle and solve for π‘₯:

\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 (π‘₯ + 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 π‘Žπ‘¦ + 𝑏π‘₯ + 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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