The Equation of a Circle

Write the equation of a circle from its centre and radius, read the centre and radius from any form, and build the equation from points or a diameter.

In the exam for IGCSE Additional Maths 0606 · 7 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What a circle equation says

A phone mast reaches every place up to 5 km away. The edge of its signal is a circle: every point on it is exactly 5 km from the mast.

Rule

A circle is every point that is the same distance from one fixed point. The fixed point is the centre. The distance is the radius.

Put the centre at $(a, b)$ and take any point $(x, y)$ on the circle. The gap across is $x - a$ and the gap up is $y - b$ . These two gaps and the radius make a right-angled triangle.

$across^{2} + up^{2} = radius^{2}$ — Pythagoras in that triangle.

$(x - a)^{2} + (y - b)^{2} = r^{2}$ — put in the two gaps and the radius r.

Formula

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

$(a, b)$ — the centre of the circle.
$r$ — the radius. The right-hand side is r squared, not r.
$(x, y)$ — any point on the circle. Every point on the circle makes the equation true.

Rule

0606 and 4037 print this formula in the List of formulas on the paper. You still need to use it quickly, so know it well.

Worked example

1
Write the equation of the circle with centre $(2, - 3)$ and radius 4.
2
$a = 2, b = - 3, r = 4$ — Read the centre and radius.
3
$(x - 2)^{2} + (y - (- 3))^{2} = 4^{2}$ — Put them into the formula.
4
$(x - 2)^{2} + (y + 3)^{2} = 16$ — Minus a negative is plus, and 4 squared is 16.

Common mistake

Writing $(y - 3)^{2}$ for a centre with $b = - 3$ . Inside the bracket the sign of the centre flips: $y - (- 3) = y + 3$ .

Key idea

René Descartes joined algebra and geometry in 1637, in a book called La Géométrie. His idea of describing points by numbers is why a circle can be written as an equation at all.

Key idea

A circle equation is just Pythagoras: across squared plus up squared equals radius squared.

Read the centre and radius

Often the paper gives the equation and asks for the centre and radius. Read them straight off, but watch the signs.

Steps

1
1. Centre: take the number inside each bracket and change its sign.
2. Radius: take the square root of the number on the right.
3. If $x^{2}$ appears with no bracket, the centre has $a = 0$ . The same goes for $y^{2}$ and $b$ .

Worked example

1
Find the centre and radius of $(x + 5)^{2} + (y - 1)^{2} = 12$ .
2
$x + 5 = x - (- 5)$ — So a = −5.
3
$y - 1$ — So b = 1.
4
$Centre (- 5, 1)$ — Signs flipped from the brackets.
5
$r = 12 = 23$ — Square root of 12, simplified. Keep it exact.

You can also test any point. Put its x and y into the left-hand side and compare the answer with $r^{2}$ .

Left side equals $r^{2}$

The point is on the circle.

Left side is less or more

Less than $r^{2}$ : inside.

More than $r^{2}$ : outside.

Worked example

1
The circle $(x - 1)^{2} + (y + 2)^{2} = 10$ . Is $(4, - 1)$ on the circle? Where is $(3, 0)$ ?
2
$(4 - 1)^{2} + (- 1 + 2)^{2} = 9 + 1 = 10$ — Equal to 10, so (4, −1) lies on the circle.
3
$(3 - 1)^{2} + (0 + 2)^{2} = 4 + 4 = 8$ — 8 is less than 10, so (3, 0) is inside the circle.

Common mistake

Giving the radius as 12 for $(x + 5)^{2} + (y - 1)^{2} = 12$ . The 12 is $r^{2}$ , so $r = 12$ .

Rule

0606 Paper 1 has no calculator. Leave a radius as an exact surd such as $23$ .

Key idea

Flip the signs for the centre. Square root the right-hand side for the radius.

When the brackets are expanded

Sometimes the equation comes multiplied out, like $x^{2} + y^{2} - 6 x + 10 y + 18 = 0$ . You cannot see the centre yet. Complete the square on the x terms and on the y terms to bring the brackets back.

Remember: $x^{2} + p x = (x + \frac{p}{2})^{2} - (\frac{p}{2})^{2}$ . Half the number, then take its square away.

Worked example

1
Find the centre and radius of $x^{2} + y^{2} - 6 x + 10 y + 18 = 0$ .
2
$x^{2} - 6 x = (x - 3)^{2} - 9$ — Half of −6 is −3, and 3 squared is 9.
3
$y^{2} + 10 y = (y + 5)^{2} - 25$ — Half of 10 is 5, and 5 squared is 25.
4
$(x - 3)^{2} - 9 + (y + 5)^{2} - 25 + 18 = 0$ — Put both back into the equation.
5
$(x - 3)^{2} + (y + 5)^{2} = 16$ — Move the numbers to the right: 9 + 25 − 18 = 16.
6
$Centre (3, - 5), r = 4$ — Read them off as before.

If you do this every time, a pattern appears. You can learn it as a shortcut. It is not on the formula list.

Formula

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

$g$ — half the number in front of x.
$f$ — half the number in front of y.
$c$ — the number on its own.

Formula

centre = (- g, - f)

$- g, - f$ — change the signs of the two halves to get the centre.

Formula

r = g^{2} + f^{2} - c

$r$ — the radius. Square both halves, add them, then take away c.
$g^{2} + f^{2} - c$ — must be more than 0. If it is not, the equation is not a real circle.

Worked example

1
Find the centre and radius of $2 x^{2} + 2 y^{2} + 8 x - 4 y - 22 = 0$ .
2
$x^{2} + y^{2} + 4 x - 2 y - 11 = 0$ — Divide every term by 2, so x² and y² each have 1 in front.
3
$g = 2, f = - 1, c = - 11$ — Half of 4 is 2, half of −2 is −1.
4
$Centre (- 2, 1)$ — Change the signs of g and f.
5
$r = 2^{2} + (- 1)^{2} - (- 11) = 16 = 4$ — 4 + 1 + 11 = 16.

Common mistake

Using the shortcut before dividing. It only works when $x^{2}$ and $y^{2}$ each have 1 in front.

Common mistake

Forgetting that c is subtracted. With $c = - 11$ you get $+ 11$ under the root.

Key idea

Expanded form hides the centre. Complete the square, or halve and flip the signs, and it appears.

Build the equation from points

Most exam questions give you facts, not the equation. You always need two things: the centre and $r^{2}$ .

If you know the centre and one point on the circle, the radius is the distance between them. Distance is Pythagoras again.

Worked example

1
A circle has centre $(- 1, 2)$ and passes through $(3, 5)$ . Find its equation.
2
$r^{2} = (3 - (- 1))^{2} + (5 - 2)^{2}$ — Distance squared from the centre to the point.
3
$r^{2} = 16 + 9 = 25$ — 4 squared plus 3 squared.
4
$(x + 1)^{2} + (y - 2)^{2} = 25$ — Centre and r squared into the formula.

If you know the two ends of a diameter, the centre is halfway between them: the midpoint. The radius is half the diameter.

Worked example

1
$A (- 4, 1)$ and $B (6, - 5)$ are the ends of a diameter. Find the equation of the circle.
2
$(\frac{- 4 + 6}{2}, \frac{1 + ( - 5 )}{2}) = (1, - 2)$ — The centre is the midpoint of AB.
3
$r^{2} = (6 - 1)^{2} + (- 5 - (- 2))^{2} = 25 + 9 = 34$ — From the centre to B. You could use A instead and get the same.
4
$(x - 1)^{2} + (y + 2)^{2} = 34$ — Leave 34 as it is. No square root needed.
5
$x^{2} + y^{2} - 2 x + 4 y - 29 = 0$ — Expanded form, if the question asks for it: 1 + 4 − 34 = −29.

The mark scheme gives a mark for the centre, a mark for the radius or $r^{2}$ , then the last mark for the equation. So show the centre and $r^{2}$ as separate lines.

Common mistake

Using the full length of AB as the radius. AB is the diameter. Find $r^{2}$ from the centre to one end.

Common mistake

Working out $r = 34 \approx 5.83$ , then squaring the rounded number. The equation needs $r^{2}$ , so keep 34 exactly.

Key idea

Centre first, then $r^{2}$ . With those two, the equation writes itself.

Clues: right angles and touching lines

Some questions hide the centre in a geometry clue. Two clues come up often.

Rule

Clue 1: an angle in a semicircle is $9 0^{\circ}$ . So if a triangle inside the circle has a right angle, the side opposite it is a diameter.

To find the right angle with coordinates, use gradients. Two lines are perpendicular when their gradients multiply to $- 1$ .

Worked example

1
The points $A (- 2, 1)$ , $B (2, 5)$ and $C (6, 1)$ lie on a circle. Find its equation.
2
$m_{A B} = \frac{5 - 1}{2 - ( - 2 )} = 1,$ $m_{B C} = \frac{1 - 5}{6 - 2} = - 1$ — Gradients of the two sides that meet at B.
3
$1 \times (- 1) = - 1$ — So angle ABC is 90°, and AC is a diameter.
4
$Centre = (\frac{- 2 + 6}{2}, \frac{1 + 1}{2}) = (2, 1)$ — Midpoint of the diameter AC.
5
$r^{2} = (6 - 2)^{2} + (1 - 1)^{2} = 16$ — From the centre to C.
6
$(x - 2)^{2} + (y - 1)^{2} = 16$ — The equation.

Rule

Clue 2: if a circle touches a line, the radius to that point meets the line at $9 0^{\circ}$ . So the distance from the centre to the line is the radius.

Worked example

1
The lines $x = 1$ , $x = 7$ , $y = - 2$ and $y = 4$ all touch a circle. Find its equation.
2
$Centre x = \frac{1 + 7}{2} = 4$ — The centre is halfway between the two upright lines.
3
$Centre y = \frac{- 2 + 4}{2} = 1$ — And halfway between the two flat lines.
4
$r = 7 - 4 = 3$ — Distance from the centre to the line x = 7.
5
$(x - 4)^{2} + (y - 1)^{2} = 9$ — The equation.

Common mistake

Taking the radius as the distance between the two lines. That distance, 6, is the diameter.

Key idea

Look for the hidden diameter: a right angle in a triangle on the circle, or two parallel lines the circle touches.

Which method? Read the question

Read what the question gives you. The wording tells you which route to take.

You see	You do
Centre and radius given	Put them straight into $(x - a)^{2} + (y - b)^{2} = r^{2}$
“Find the centre and radius” of $(x - a)^{2} + (y - b)^{2} = k$	Flip the signs, square root k
$x^{2} + y^{2} + \dots = 0$	Complete the square, or use $(- g, - f)$ and $g^{2} + f^{2} - c$
Centre and a point on the circle	$r^{2}$ = distance squared
Ends of a diameter	Midpoint for the centre, then $r^{2}$ to one end
Three points with a right angle	The side opposite the right angle is a diameter
Lines that touch the circle	Centre halfway between parallel lines, radius to a line
“Show that the point lies on the circle”	Substitute and show the left side equals $r^{2}$

Common mistake

Leaving the answer as $r = 5$ when the question asks for the equation. Finish by writing the full equation.

Key idea

Every circle question needs two things: the centre and $r^{2}$ . Find those, and you are done.

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Real past-paper questions on the equation of a circle

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