Equation of a Circle

What do I need to know about the equation of a circle?

A circle with centre (a, b) and radius r has the equation

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

Eqn of Circle graph illustr, A Level & AS Level Pure Maths Revision Notes

You need to be able to find the centre and radius of a circle from its equation

Eqn of Circle Illustr 1, A Level & AS Level Pure Maths Revision Notes

You need to be able to find the equation of a circle given its centre and radius

Eqn of Circle Illustr 2, A Level & AS Level Pure Maths Revision Notes

Exam Tip

Remember that the numbers in the brackets have the opposite signs to the coordinates of the centre.
Don't forget to take the square root of the right-hand side of the equation when finding the radius.

Worked example

Eqn of Circle Example, A Level & AS Level Pure Maths Revision Notes

Intersection of a Circle and a Line

How can a circle and a straight line intersect?

A circle and a straight line could
- intersect at two distinct (different) points
- intersect at one point only
  - in this case the line would be tangent to the circle
- not intersect at all

equation-of-a-circle-rn-image

How do I find the intersections between a line and a circle?

As with two straight lines, this is a matter of solving the two equations simultaneously
- However the equation for the circle is quadratic so the method involves more steps
First, ensure that the equation for the circle is in the form ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$
- Where the centre of the circle is (a, b) and its radius is r
Also ensure the equation for the straight line is in the form $y = m x + c$
- m is the gradient and c is the y-axis intercept
One or both equations may need deducing from a sketch or information given in a question
- They may need to be rearranged to get them in the correct format

STEP 1
Substitute the equation of the straight line into the equation of the circle for

y

e.g.

For the circle

{(x - 2)}^{2} + {(y - 3)}^{2} = 2^{2}

and straight line

y = x - 1

substituting the line into the circle would give

{(x - 2)}^{2} + {((x - 1) - 3)}^{2} = 2^{2}

STEP 2

Expand and simplify to give a quadratic equation in

x

Quadratic equations are easiest to deal with once in the form "

a x^{2} + b x + c = 0

e.g.

\begin{array}{rcl} {(x - 2)}^{2} + {(x - 4)}^{2} & = & 4 \\ x^{2} - 4 x + 4 + x^{2} - 8 x + 16 - 4 & = & 0 \\ 2 x^{2} - 12 x + 16 & = & 0 \\ x^{2} - 6 x + 8 & = & 0 \end{array}

STEP 3
Solve the quadratic equation to find the x-coordinates of the points of intersection

e.g.

\begin{array}{rcl} (x - 2) (x - 4) & = & 0 \\ x & = & 2 \\ x & = & 4 \end{array}

STEP 4
Substitute the x-coordinates into the equation of the straight line to find the y-coordinates

(You could use the circle equation here but that's normally harder!)

e.g.

\begin{array}{rcl} y & = & 2 - 1 = 1 \\ y & = & 4 - 1 = 3 \end{array}

STEP 5

Write your final answer as coordinate pairs

e.g.

(2, 1)

and

(4, 3)

The number of solutions to the quadratic equation will determine the number of intersections between the line and the equation (or vice versa)
- a quadratic with two (real) solutions will indicate two intersections
- a quadratic with only one (real) solution (or a repeated (real) solution) will indicate one intersection
  - so the line would be a tangent to the circle
- a quadratic with no (real) solutions will indicate the line and circle do not intersect

Worked example

An equation with centre (-3, 2) and radius 4 intersects the straight line with equation $y = x + 1$ at two points, A and B.
Find the coordinates of points A and B.

Write the equation of the circle using the information given, the centre is (-3, 2) and the radius is 4

${(x + 3)}^{2} + {(y - 2)}^{2} = 4^{2} {(x + 3)}^{2} + {(y - 2)}^{2} = 16$

Substitute the equation of the line, $y = x + 1$ , for $y$ , and simplify

$\begin{array}{rcl} {(x + 3)}^{2} + {((x + 1) - 2)}^{2} & = & 16 \\ {(x + 3)}^{2} + {(x - 1)}^{2} & = & 16 \end{array}$

Expand the brackets, simplify, and equate to zero

$\begin{array}{rcl} x^{2} + 6 x + 9 + x^{2} - 2 x + 1 & = & 16 \\ 2 x^{2} + 4 x + 10 & = & 16 \\ 2 x^{2} + 4 x - 6 & = & 0 \end{array}$

We now need to solve the quadratic
As all the coefficients are even, we can divide both sides by 2 to make this easier

$x^{2} + 2 x - 3 = 0$

Factorise and solve for $x$

$(x + 3) (x - 1) = 0 x = - 3, x = 1$

These are the $x$ -coordinates of the points of intersection of the line and the circle
To find the $y$ -coordinates, substitute these into the equation of the line; $y = x + 1$

$x = - 3 : y = (- 3) + 1 = - 2 x = 1 : y = (1) + 1 = 2$

Write the coordinate pairs as your final answer
Be careful to match the corresponding $x$ and $y$ values correctly

$(- 3, - 2) (1, 2)$

Equation of a Circle (AQA GCSE Further Maths)

Revision Note