The equation of a circle

AS · P1 · 15 min

A circle is the set of all points at a fixed distance, the radius, from a fixed point, the centre. Write that sentence using the distance formula and you have the equation of a circle. Paper 1 asks you to read the centre and radius from an equation (including the expanded form), to build the equation from given facts, and to use the classic circle theorems, such as the tangent being perpendicular to the radius, with coordinates. How a line meets a circle has its own note, Lines and circles.

From the definition to the equation

Let the centre be $C(a, b)$ and the radius $r$ . A point $P(x, y)$ is on the circle exactly when $CP = r$ . By the distance formula,

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

Squaring both sides gives the standard equation.

Key result

The circle with centre $(a, b)$ and radius $r$ has equation

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

A circle with centre at the origin is $x^2 + y^2 = r^2$ .

The signs inside the brackets are the opposite of the centre's coordinates: $(x + 2)^2 + (y - 5)^2 = 9$ has centre $(-2, 5)$ and radius $3$ . And the right-hand side is $r^2$ , not $r$ : $(x - 1)^2 + y^2 = 20$ has radius $\sqrt{20} = 2\sqrt{5}$ .

(x - 2)^2 + (y + 1)^2 = 16 (2, -1) (2, -1) -- (6, -1)

The circle $(x - 2)^2 + (y + 1)^2 = 16$ : centre $(2, -1)$ , radius $4$ .

Inside, on or outside

The same distance tells you where any point is. Compute $d^2 = (x - a)^2 + (y - b)^2$ for the point and compare with $r^2$ : if $d^2 < r^2$ the point is inside, if $d^2 = r^2$ it is on the circle, and if $d^2 > r^2$ it is outside.

The expanded form

Multiplying out $(x - a)^2 + (y - b)^2 = r^2$ gives an equation of the shape

x^2 + y^2 + 2gx + 2fy + c = 0

Recognise a circle by two features: the coefficients of $x^2$ and $y^2$ are equal, and there is no $xy$ term. (If the $x^2$ and $y^2$ coefficients are equal but not $1$ , divide through first.)

To find the centre and radius, complete the square in $x$ and in $y$ separately (see Completing the square):

(x + g)^2 - g^2 + (y + f)^2 - f^2 + c = 0 \quad\Rightarrow\quad (x + g)^2 + (y + f)^2 = g^2 + f^2 - c

Key result

$x^2 + y^2 + 2gx + 2fy + c = 0$ has

\text{centre } (-g,\ -f), \qquad \text{radius } \sqrt{g^2 + f^2 - c}

It is a real circle only if $g^2 + f^2 - c > 0$ .

In practice, completing the square every time is safer than remembering the formula, and it is what the mark scheme rewards. Note that $2g$ is the coefficient of $x$ , so $g$ is half of it.

Centre and radius from the expanded form

If needed, divide so the coefficients of $x^2$ and $y^2$ are $1$ .
Group the $x$ terms and the $y$ terms; move the constant to the right.
Complete the square in $x$ and in $y$ , adding the squared halves to the right-hand side.
Read off the centre and the radius (the square root of the right-hand side).

Geometry you are expected to use

The syllabus names three properties of circles to use in coordinate problems.

Key result

Tangent and radius. The tangent at a point is perpendicular to the radius drawn to that point.
Angle in a semicircle. If $AB$ is a diameter and $P$ is any other point on the circle, angle $APB = 90^\circ$ . Conversely, if angle $APB = 90^\circ$ , then $P$ lies on the circle with diameter $AB$ .
Symmetry. A circle is symmetrical about every line through its centre. In particular, the perpendicular bisector of any chord passes through the centre.

Each gives a standard technique:

The tangent at $P$ : find the gradient of the radius $CP$ , then use the perpendicular gradient through $P$ . The normal at $P$ is the line $CP$ itself.
Right angle in a triangle: the hypotenuse is a diameter, so its midpoint is the centre of the circle through all three vertices.
Two points on the circle: the centre lies on the perpendicular bisector of the chord joining them. Two chords give two bisectors, which meet at the centre.

(x - 2)^2 + (y + 1)^2 = 16 (2, -1) -- (2 + 16/5, -1 + 12/5) y = -(4/3)*(x - 26/5) + 7/5

The radius to the point $\left(\tfrac{26}{5}, \tfrac{7}{5}\right)$ has gradient $\tfrac{3}{4}$ , so the tangent there has gradient $-\tfrac{4}{3}$ .

Finding the equation of a circle

You need the centre and the radius. Questions give them indirectly:

Information given	How to get the centre and radius
ends of a diameter $A$ , $B$	centre $=$ midpoint of $AB$ , $r = \tfrac{1}{2}AB$
centre and a point on the circle	$r =$ distance from centre to the point
circle touches the $x$ -axis	$r =
circle touches the $y$ -axis	$r =
two points and a line containing the centre	perpendicular bisector of the two points meets the line at the centre
three points	either two perpendicular bisectors, or substitute the points into $x^2 + y^2 + 2gx + 2fy + c = 0$ and solve for $g$ , $f$ , $c$
right-angled triangle inscribed	the hypotenuse is a diameter

Worked examples

Centre and radius from the expanded form

Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ .

Solution

\begin{aligned} (x^2 - 6x) + (y^2 + 4y) &= 12 \\ (x - 3)^2 - 9 + (y + 2)^2 - 4 &= 12 \\ (x - 3)^2 + (y + 2)^2 &= 25 \end{aligned}

Centre $(3, -2)$ , radius $5$ .

A circle from a diameter

The points $A(-1, 4)$ and $B(5, -4)$ are the ends of a diameter of a circle. Find the equation of the circle in the form $x^2 + y^2 + 2gx + 2fy + c = 0$ .

Solution

Centre $=$ midpoint of $AB = \left(\tfrac{-1 + 5}{2}, \tfrac{4 - 4}{2}\right) = (2, 0)$ .

$AB = \sqrt{6^2 + 8^2} = 10$ , so $r = 5$ .

(x - 2)^2 + y^2 = 25 \quad\Rightarrow\quad x^2 - 4x + 4 + y^2 = 25 \quad\Rightarrow\quad x^2 + y^2 - 4x - 21 = 0

Check: $A$ gives $1 + 16 + 4 - 21 = 0$ . Correct.

The tangent at a point

Show that $P(3, 2)$ lies on the circle $(x - 1)^2 + (y + 2)^2 = 20$ , and find the equation of the tangent at $P$ .

Solution

$(3 - 1)^2 + (2 + 2)^2 = 4 + 16 = 20$ , so $P$ is on the circle.

The centre is $C(1, -2)$ . The radius $CP$ has gradient $\dfrac{2 - (-2)}{3 - 1} = 2$ . The tangent is perpendicular to the radius, so its gradient is $-\tfrac{1}{2}$ :

y - 2 = -\tfrac{1}{2}(x - 3) \quad\Rightarrow\quad 2y - 4 = -x + 3 \quad\Rightarrow\quad x + 2y = 7

Centre on a given line

A circle passes through $P(1, 6)$ and $Q(7, 4)$ , and its centre lies on the line $y = x - 1$ . Find the equation of the circle.

Solution

The centre is on the perpendicular bisector of the chord $PQ$ . Midpoint $(4, 5)$ ; $m_{PQ} = \dfrac{4 - 6}{7 - 1} = -\dfrac{1}{3}$ , so the bisector has gradient $3$ :

y - 5 = 3(x - 4) \quad\Rightarrow\quad y = 3x - 7

It meets $y = x - 1$ where $3x - 7 = x - 1$ , so $x = 3$ , $y = 2$ . The centre is $(3, 2)$ .

r^2 = (1 - 3)^2 + (6 - 2)^2 = 4 + 16 = 20

The circle is $(x - 3)^2 + (y - 2)^2 = 20$ .

A circle through three points

Find the equation of the circle through $A(5, 3)$ , $B(-2, 2)$ and $C(6, -4)$ , and state its centre and radius.

Solution

Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ . Each point satisfies it:

\begin{aligned} A:&\quad 34 + 10g + 6f + c = 0 \\ B:&\quad 8 - 4g + 4f + c = 0 \\ C:&\quad 52 + 12g - 8f + c = 0 \end{aligned}

Subtract to remove $c$ :

\begin{aligned} A - B:&\quad 26 + 14g + 2f = 0 \quad\Rightarrow\quad 7g + f = -13 \\ C - B:&\quad 44 + 16g - 12f = 0 \quad\Rightarrow\quad 4g - 3f = -11 \end{aligned}

From the first, $f = -13 - 7g$ . Then $4g + 39 + 21g = -11$ , so $25g = -50$ , $g = -2$ and $f = 1$ . From $B$ : $c = -8 + 4g - 4f = -8 - 8 - 4 = -20$ .

x^2 + y^2 - 4x + 2y - 20 = 0

Centre $(-g, -f) = (2, -1)$ , radius $\sqrt{4 + 1 + 20} = 5$ .

Touching both axes

A circle lies in the first quadrant and touches both the $x$ -axis and the $y$ -axis. It passes through the point $(3, 6)$ . Find the two possible equations.

Solution

A circle touching both axes in the first quadrant has its centre at the same distance $a$ from each axis: centre $(a, a)$ , radius $a$ . Substituting $(3, 6)$ :

\begin{aligned} (3 - a)^2 + (6 - a)^2 &= a^2 \\ 9 - 6a + a^2 + 36 - 12a + a^2 &= a^2 \\ a^2 - 18a + 45 &= 0 \\ (a - 3)(a - 15) &= 0 \end{aligned}

So the circle is $(x - 3)^2 + (y - 3)^2 = 9$ or $(x - 15)^2 + (y - 15)^2 = 225$ .

A tangent, an intercept and an area

The circle $x^2 + y^2 - 4x + 6y - 12 = 0$ has centre $C$ . The point $P(6, 0)$ lies on the circle.

(a) Find the centre and radius.

(b) The tangent at $P$ meets the $y$ -axis at $T$ . Find $T$ .

Solution

(a) $(x - 2)^2 - 4 + (y + 3)^2 - 9 = 12$ , so $(x - 2)^2 + (y + 3)^2 = 25$ . Centre $C(2, -3)$ , radius $5$ .

(b) $m_{CP} = \dfrac{0 - (-3)}{6 - 2} = \dfrac{3}{4}$ , so the tangent has gradient $-\dfrac{4}{3}$ : $y = -\tfrac{4}{3}(x - 6)$ . At $x = 0$ , $y = 8$ , so $T = (0, 8)$ .

(c) The angle at $P$ is $90^\circ$ (tangent perpendicular to radius), so $CP$ and $PT$ are the base and height. $CP = 5$ and $PT = \sqrt{6^2 + 8^2} = 10$ .

\text{Area} = \tfrac{1}{2} \times 5 \times 10 = 25

Check with Pythagoras: $CT^2 = 4 + 121 = 125 = 25 + 100$ .

Watch out

Wrong signs for the centre. $(x + 3)^2 + (y - 1)^2 = 16$ has centre $(-3, 1)$ , not $(3, -1)$ .

Forgetting the square root. $(x - 2)^2 + y^2 = 20$ has radius $\sqrt{20}$ , not $20$ .

Completing the square on one side only. When you add $9$ and $4$ to complete the squares, add them to the other side too (or subtract them on the same side).

Not checking it is a circle. $x^2 + y^2 - 2x + 6y + 15 = 0$ gives $(x - 1)^2 + (y + 3)^2 = -5$ , which has no points. If asked to "show that the equation represents a circle", show the right-hand side is positive.

Tangent through the centre. The tangent passes through the point of contact $P$ , with gradient perpendicular to $CP$ . The line through the centre is the normal.

Exam tip

Form of the answer. Unless a form is specified, $(x - a)^2 + (y - b)^2 = r^2$ is accepted. If the question asks for $x^2 + y^2 + ax + by + c = 0$ , expand.
State the property you use. Write "the tangent is perpendicular to the radius" or "the angle in a semicircle is $90^\circ$ ". It shows the examiner your method and often earns the method mark.
Show a point lies on a circle by substituting it and showing both sides agree, not by drawing.
Exact answers. Leave the radius as a surd, such as $2\sqrt{5}$ .
Long questions. A circle question is typically 7 to 10 marks: centre and radius, a tangent, then an intersection with a line. Keep exact values throughout.

Summary

$(x - a)^2 + (y - b)^2 = r^2$ : centre $(a, b)$ , radius $r$ .
$x^2 + y^2 + 2gx + 2fy + c = 0$ : complete the square; centre $(-g, -f)$ , radius $\sqrt{g^2 + f^2 - c}$ , provided this is positive.
Point inside, on or outside: compare its squared distance from the centre with $r^2$ .
Tangent at $P$ : perpendicular to the radius $CP$ , through $P$ .
A right angle subtended by $AB$ means $AB$ is a diameter.
The centre lies on the perpendicular bisector of every chord.
Three points: two perpendicular bisectors, or three equations in $g$ , $f$ , $c$ .

Practice questions

Question

Find the centre and radius of the circle $x^2 + y^2 + 8x - 2y + 8 = 0$ .
Find the equation of the circle with centre $(-2, 5)$ that touches the $x$ -axis.
Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(-3, 4)$ .
Show that the equation $x^2 + y^2 - 2x + 6y + 15 = 0$ does not represent a circle.
A circle has centre $(4, 1)$ and passes through $(7, 5)$ . Determine whether each of the points $(0, -1)$ and $(1, 6)$ lies inside, on or outside the circle.
Find the equation of the circle passing through $(0, 0)$ , $(8, 0)$ and $(0, 6)$ .
A circle passes through $(2, 1)$ and $(6, 5)$ and has its centre on the $y$ -axis. Find its equation.
The circle $(x - 3)^2 + (y + 1)^2 = 10$ has centre $C$ and passes through $P(4, 2)$ . The tangent at $P$ meets the $x$ -axis at $T$ . Find the equation of the tangent, the coordinates of $T$ and the area of triangle $CPT$ .
A circle touches the $y$ -axis, has its centre on the line $y = 3$ , and passes through $(2, 7)$ . Find its equation.
The circle $x^2 + y^2 - 6x - 2y - 15 = 0$ passes through $A(0, 5)$ . (a) Find the coordinates of $B$ , the other end of the diameter through $A$ . (b) Show that $D(7, 4)$ lies on the circle and that angle $ADB = 90^\circ$ . (c) Find the equations of the tangents at $A$ and at $B$ , and explain why they are parallel.

Answers

$(x + 4)^2 - 16 + (y - 1)^2 - 1 + 8 = 0$ , so $(x + 4)^2 + (y - 1)^2 = 9$ . Centre $(-4, 1)$ , radius $3$ .
The radius is the distance from the centre to the $x$ -axis, which is $5$ : $(x + 2)^2 + (y - 5)^2 = 25$ .
The radius from $(0, 0)$ to $(-3, 4)$ has gradient $-\tfrac{4}{3}$ , so the tangent has gradient $\tfrac{3}{4}$ : $y - 4 = \tfrac{3}{4}(x + 3)$ , i.e. $3x - 4y + 25 = 0$ .
$(x - 1)^2 - 1 + (y + 3)^2 - 9 + 15 = 0$ gives $(x - 1)^2 + (y + 3)^2 = -5$ . A sum of squares cannot be negative, so no points satisfy it: it is not a circle.
$r^2 = 3^2 + 4^2 = 25$ . For $(0, -1)$ : $d^2 = 16 + 4 = 20 < 25$ , inside. For $(1, 6)$ : $d^2 = 9 + 25 = 34 > 25$ , outside.
The axes meet at right angles at $(0, 0)$ , so the chord from $(8, 0)$ to $(0, 6)$ subtends $90^\circ$ there and is a diameter. Centre $(4, 3)$ , radius $\tfrac{1}{2}\sqrt{64 + 36} = 5$ : $(x - 4)^2 + (y - 3)^2 = 25$ , or $x^2 + y^2 - 8x - 6y = 0$ .
Midpoint $(4, 3)$ ; chord gradient $1$ , so the perpendicular bisector is $y - 3 = -(x - 4)$ , i.e. $y = 7 - x$ . On the $y$ -axis, $x = 0$ : centre $(0, 7)$ . $r^2 = 2^2 + (1 - 7)^2 = 40$ . Circle: $x^2 + (y - 7)^2 = 40$ .
$C = (3, -1)$ ; $m_{CP} = \tfrac{3}{1} = 3$ , so the tangent has gradient $-\tfrac{1}{3}$ : $y - 2 = -\tfrac{1}{3}(x - 4)$ , i.e. $x + 3y = 10$ . At $y = 0$ , $T = (10, 0)$ . $CP = \sqrt{10}$ and $PT = \sqrt{36 + 4} = \sqrt{40}$ , with a right angle at $P$ . Area $= \tfrac{1}{2}\sqrt{10}\sqrt{40} = \tfrac{1}{2}(20) = 10$ .
Centre $(a, 3)$ ; touching the $y$ -axis means $r = |a|$ . Then $(2 - a)^2 + (7 - 3)^2 = a^2$ , so $4 - 4a + 16 = 0$ and $a = 5$ . Circle: $(x - 5)^2 + (y - 3)^2 = 25$ .
(a) $(x - 3)^2 + (y - 1)^2 = 25$ , centre $(3, 1)$ . The centre is the midpoint of $AB$ , so $B = (6, -3)$ . (b) $(7 - 3)^2 + (4 - 1)^2 = 16 + 9 = 25$ , so $D$ is on the circle. $m_{AD} = \dfrac{4 - 5}{7 - 0} = -\dfrac{1}{7}$ and $m_{BD} = \dfrac{4 + 3}{7 - 6} = 7$ ; the product is $-1$ , so angle $ADB = 90^\circ$ (as expected for an angle in a semicircle). (c) The radius to $A$ has gradient $\dfrac{5 - 1}{0 - 3} = -\dfrac{4}{3}$ , so the tangent at $A$ has gradient $\tfrac{3}{4}$ : $y - 5 = \tfrac{3}{4}x$ , i.e. $3x - 4y + 20 = 0$ . The radius to $B$ lies on the same diameter, gradient $-\tfrac{4}{3}$ , so the tangent at $B$ also has gradient $\tfrac{3}{4}$ : $y + 3 = \tfrac{3}{4}(x - 6)$ , i.e. $3x - 4y - 30 = 0$ . Both tangents are perpendicular to the same diameter, so they are parallel.

From the definition to the equation

Inside, on or outside

The expanded form

Geometry you are expected to use

Finding the equation of a circle

Worked examples

Practice questions

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