Circles

What you'll learn

Write the equation of a circle from its centre, radius, or diameter.
Complete the square to find the centre and radius from an expanded equation.
Find intersections between circles and straight lines.
Use perpendicular gradients to find equations of tangents.

Prerequisites: coordinate-geometry tools

Before circles, you need three tools: distance, midpoint, and gradient.

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$ :

A coordinate diagram showing how distance, midpoint and gradient relate to the same line segment between two points.

Distance measures the length of the line segment joining the points:
$AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
The midpoint is the point halfway between them:
$(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
Gradient measures steepness:
$m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}$

Example

Using distance, midpoint and gradient

For $A (- 2, 4)$ and $B (6, - 2)$ :

Find the distance using the distance formula.
$AB=(6−(−2))2+(−2−4)2=82+(−6)2=100=10AB=\sqrt{(6-(-2))^2+(-2-4)^2} =\sqrt{8^2+(-6)^2} =\sqrt{100}=10$
Find the midpoint by averaging the coordinates.
$(−2+62,4+(−2)2)=(2,1)\left(\frac{-2+6}{2},\frac{4+(-2)}{2}\right)=(2,1)$
Find the gradient.
$m=−2−46−(−2)=−68=−34m=\frac{-2-4}{6-(-2)}=\frac{-6}{8}=-\frac{3}{4}$

The standard equation of a circle

Definition

Circle

A circle is the set of all points that are the same distance from one fixed point. The fixed point is the centre, and the fixed distance is the radius.

If a circle has centre $(a, b)$ and radius $r$ , its equation is:

A labelled coordinate diagram showing the centre, radius and a general point on a circle in standard form.

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

Notice the signs: centre $(a, b)$ gives brackets $(x - a)$ and $(y - b)$ .

A labelled coordinate diagram showing the centre, radius, and standard equation of a circle

Key Idea

Standard form

The equation $x-a)^2+(y-b)^2=r^2$ means: “the distance from any point $(x, y)$ on the circle to the centre $(a, b)$ is always $r$ .”

Example

Finding the equation from a centre and a point

A circle has centre $(3, - 1)$ and passes through $P (7, 2)$ . Find its equation.

The radius is found from the right-angled triangle between the centre and the point on the circle.

Use the centre as $(a, b) = (3, - 1)$ , so the equation begins:
$x-3)^2+(y+1)^2=r^2$
Find $r^2$ using the distance from the centre to $P$ .
$r^2=(7-3)^2+(2-(-1))^2=4^2+3^2=25$
Substitute into the circle equation.
$x-3)^2+(y+1)^2=25$

Circles from a diameter

If you are told that a line segment is a diameter, its endpoints lie opposite each other on the circle. The centre is the midpoint of the diameter.

Definition

Diameter

A diameter is a chord that passes through the centre of the circle. Its length is twice the radius.

Example

Finding a circle from the endpoints of a diameter

The points $A (- 3, 5)$ and $B (9, - 1)$ are the endpoints of a diameter. Find the equation of the circle.

The midpoint of the diameter gives the centre, and half the diameter gives the radius.

Find the centre using the midpoint of $A B$ .
$(−3+92,5+(−1)2)=(3,2)\left(\frac{-3+9}{2},\frac{5+(-1)}{2}\right)=(3,2)$
Find $r^2$ using the distance from the centre to one endpoint.
$r^2=(-3-3)^2+(5-2)^2=(-6)^2+3^2=45$
Write the equation.
$x-3)^2+(y-2)^2=45$

Tip

Right-angle shortcut

If three points on a circle make a right angle at one point, the side opposite the right angle is a diameter. This is often the fastest way to locate the centre.

A right angle subtended by a diameter shows that the hypotenuse of the triangle is the circle’s diameter.

Completing the square

Circle equations are not always given in standard form. You may see something like:

x^2+y^2+4x-10y=7

To find the centre and radius, you complete the square separately for the $x$ terms and the $y$ terms.

Example

Finding the centre and radius by completing the square

Find the centre and radius of the circle

x^2+y^2+4x-10y=7

Group the $x$ terms and $y$ terms.
$x^2+4x+y^2-10y=7$
Complete the square for each pair.
$x^2+4x=(x+2)^2-4$ $y^2-10y=(y-5)^2-25$
Substitute these into the equation.
$x+2)^2-4+(y-5)^2-25=7$
Move the constants to the right-hand side.
$x+2)^2+(y-5)^2=36$
Read off the centre and radius.
$=(−2,5),r=6\text{centre }=(-2,5),\quad r=6$

Common Mistake

Sign errors in the centre

If the equation contains $x+2)^2$ , the centre has $x$ -coordinate -2, not 2. The sign inside the bracket is the opposite sign of the centre coordinate.

Finding a missing constant

If a point lies on a circle, its coordinates satisfy the circle equation. Substitute the point in to find the missing value.

Example

Using a point on the circle to find a constant

The circle has equation

x^2+y^2-6x+4y+k=0

and passes through $(2, 3)$ . Find $k$ , then find the centre and radius.

Substitute $x = 2$ and $y = 3$ into the equation.
$2^2+3^2-6(2)+4(3)+k=0$
Simplify to find $k$ .
$4 + 9 - 12 + 12 + k = 0$ $13 + k = 0$ $k = - 13$
Write the equation with $k = - 13$ .
$x^2+y^2-6x+4y-13=0$
Complete the square.
$x-3)^2-9+(y+2)^2-4-13=0$
Rearrange into standard form.
$x-3)^2+(y+2)^2=26$
Read off the centre and radius.
$=(3,−2),r=26\text{centre }=(3,-2),\quad r=\sqrt{26}$

Intersections with lines

To find where a line meets a circle, substitute the line equation into the circle equation. This usually creates a quadratic.

A line can meet a circle in two points, touch it once as a tangent, or miss it completely depending on the quadratic roots.

Two distinct roots means the line cuts the circle at two points.
One repeated root means the line is a tangent.
No real roots means the line misses the circle.

Example

Finding the points where a line meets a circle

The circle has centre $(2, 3)$ and radius 5. The line $y = x + 2$ intersects the circle at $A$ and $B$ . Find their coordinates.

The line crosses the circle at the two points found by solving the resulting quadratic.

Write the circle equation.
$x-2)^2+(y-3)^2=25$
Substitute $y = x + 2$ into the circle equation.
$x-2)^2+(x+2-3)^2=25$
Simplify and solve the quadratic.
$x-2)^2+(x-1)^2=25$ $x^2-4x+4+x^2-2x+1=25$ $2x^2-6x-20=0$ $x^2-3x-10=0$ $(x - 5) (x + 2) = 0$
Find the corresponding $y$ values using $y = x + 2$ .
$\Rightarrow y=7$ $\Rightarrow y=0$
State the intersection points.
$A(5,7),B(−2,0)A(5,7),\quad B(-2,0)$

Tip

Axis intersections

For intersections with the $y$ -axis, set $x = 0$ . For intersections with the $x$ -axis, set $y = 0$ .

Tangents to circles

Definition

Tangent

A tangent is a straight line that touches a circle at exactly one point. The point where it touches is called the point of contact.

The key fact is that the radius to the point of contact is perpendicular to the tangent.

A labelled diagram showing a radius meeting a tangent at right angles

If two non-vertical lines are perpendicular, their gradients multiply to -1. So if the radius has gradient $m$ , the tangent has gradient $−1m-\frac{1}{m}$ .

Example

Finding the equation of a tangent at a point

A circle has centre $(- 1, 2)$ and passes through $A (4, 4)$ . Find the equation of the tangent at $A$ in the form $a x + b y + c = 0$ .

The tangent at a point is perpendicular to the radius drawn to that point.

Find the gradient of the radius from the centre to $A$ .
$mradius=4−24−(−1)=25m_{\text{radius}}=\frac{4-2}{4-(-1)}=\frac{2}{5}$
Use the negative reciprocal for the tangent gradient.
$mtangent=−52m_{\text{tangent}}=-\frac{5}{2}$
Use the point-gradient form through $A (4, 4)$ .
$y−4=−52(x−4)y-4=-\frac{5}{2}(x-4)$
Rearrange into integer form.
$2 y - 8 = - 5 x + 20$ $5 x + 2 y - 28 = 0$

Common Mistake

Using the radius gradient as the tangent gradient

The tangent is not parallel to the radius. It is perpendicular to it, so you must use the negative reciprocal gradient.

Tangency using distance from a point to a line

Sometimes you are given a centre and a tangent line, but not the radius. The radius is the perpendicular distance from the centre to the tangent line.

For a line $a x + b y + c = 0$ , the distance from $x_1,y_1)$ to the line is:

∣ax1+by1+c∣a2+b2\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}

Example

Finding a circle from its centre and a tangent line

A circle has centre $(6, - 1)$ . The line $3 x + y - 4 = 0$ is tangent to the circle. Find the equation of the circle.

Use the distance from the centre to the tangent line to find the radius.
$r=∣3(6)+(−1)−4∣32+12r=\frac{|3(6)+(-1)-4|}{\sqrt{3^2+1^2}}$
Simplify.
$r=∣18−1−4∣10=1310r=\frac{|18-1-4|}{\sqrt{10}}=\frac{13}{\sqrt{10}}$
Square the radius for the circle equation.
$r2=(1310)2=16910r^2=\left(\frac{13}{\sqrt{10}}\right)^2=\frac{169}{10}$
Write the equation using centre $(6, - 1)$ .
$(x−6)2+(y+1)2=16910(x-6)^2+(y+1)^2=\frac{169}{10}$

Exam technique

In the exam

Put circle equations into standard form as early as possible, because the centre and radius then become easy to read.
For tangents, draw a quick sketch and mark the radius at 90° to the tangent.
When substituting a line into a circle, expect a quadratic and use the discriminant if the question asks about tangency or ranges.

Self review

Check yourself

Can you find the centre and radius of $x^2+y^2-8x+6y=12$ ?
If a circle has centre $(4, - 2)$ and passes through $(1, 2)$ , can you write its equation?
Can you find the tangent gradient at a point once you know the gradient of the radius?

Circles

What you'll learn

Write the equation of a circle from its centre, radius, or diameter.
Complete the square to find the centre and radius from an expanded equation.
Find intersections between circles and straight lines.
Use perpendicular gradients to find equations of tangents.

Prerequisites: coordinate-geometry tools

Before circles, you need three tools: distance, midpoint, and gradient.

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$ :

A coordinate diagram showing how distance, midpoint and gradient relate to the same line segment between two points.

Distance measures the length of the line segment joining the points:
$AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
The midpoint is the point halfway between them:
$(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
Gradient measures steepness:
$m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}$

Example

Using distance, midpoint and gradient

For $A (- 2, 4)$ and $B (6, - 2)$ :

Find the distance using the distance formula.
$AB=(6−(−2))2+(−2−4)2=82+(−6)2=100=10AB=\sqrt{(6-(-2))^2+(-2-4)^2} =\sqrt{8^2+(-6)^2} =\sqrt{100}=10$
Find the midpoint by averaging the coordinates.
$(−2+62,4+(−2)2)=(2,1)\left(\frac{-2+6}{2},\frac{4+(-2)}{2}\right)=(2,1)$
Find the gradient.
$m=−2−46−(−2)=−68=−34m=\frac{-2-4}{6-(-2)}=\frac{-6}{8}=-\frac{3}{4}$

The standard equation of a circle

Definition

Circle

A circle is the set of all points that are the same distance from one fixed point. The fixed point is the centre, and the fixed distance is the radius.

If a circle has centre $(a, b)$ and radius $r$ , its equation is:

A labelled coordinate diagram showing the centre, radius and a general point on a circle in standard form.

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

Notice the signs: centre $(a, b)$ gives brackets $(x - a)$ and $(y - b)$ .

A labelled coordinate diagram showing the centre, radius, and standard equation of a circle

Key Idea

Standard form

The equation $x-a)^2+(y-b)^2=r^2$ means: “the distance from any point $(x, y)$ on the circle to the centre $(a, b)$ is always $r$ .”

Example

Finding the equation from a centre and a point

A circle has centre $(3, - 1)$ and passes through $P (7, 2)$ . Find its equation.

The radius is found from the right-angled triangle between the centre and the point on the circle.

Use the centre as $(a, b) = (3, - 1)$ , so the equation begins:
$x-3)^2+(y+1)^2=r^2$
Find $r^2$ using the distance from the centre to $P$ .
$r^2=(7-3)^2+(2-(-1))^2=4^2+3^2=25$
Substitute into the circle equation.
$x-3)^2+(y+1)^2=25$

Circles from a diameter

If you are told that a line segment is a diameter, its endpoints lie opposite each other on the circle. The centre is the midpoint of the diameter.

Definition

Diameter

A diameter is a chord that passes through the centre of the circle. Its length is twice the radius.

Example

Finding a circle from the endpoints of a diameter

The points $A (- 3, 5)$ and $B (9, - 1)$ are the endpoints of a diameter. Find the equation of the circle.

The midpoint of the diameter gives the centre, and half the diameter gives the radius.

Find the centre using the midpoint of $A B$ .
$(−3+92,5+(−1)2)=(3,2)\left(\frac{-3+9}{2},\frac{5+(-1)}{2}\right)=(3,2)$
Find $r^2$ using the distance from the centre to one endpoint.
$r^2=(-3-3)^2+(5-2)^2=(-6)^2+3^2=45$
Write the equation.
$x-3)^2+(y-2)^2=45$

Tip

Right-angle shortcut

If three points on a circle make a right angle at one point, the side opposite the right angle is a diameter. This is often the fastest way to locate the centre.

A right angle subtended by a diameter shows that the hypotenuse of the triangle is the circle’s diameter.

Completing the square

Circle equations are not always given in standard form. You may see something like:

x^2+y^2+4x-10y=7

To find the centre and radius, you complete the square separately for the $x$ terms and the $y$ terms.

Example

Finding the centre and radius by completing the square

Find the centre and radius of the circle

x^2+y^2+4x-10y=7

Group the $x$ terms and $y$ terms.
$x^2+4x+y^2-10y=7$
Complete the square for each pair.
$x^2+4x=(x+2)^2-4$ $y^2-10y=(y-5)^2-25$
Substitute these into the equation.
$x+2)^2-4+(y-5)^2-25=7$
Move the constants to the right-hand side.
$x+2)^2+(y-5)^2=36$
Read off the centre and radius.
$=(−2,5),r=6\text{centre }=(-2,5),\quad r=6$

Common Mistake

Sign errors in the centre

If the equation contains $x+2)^2$ , the centre has $x$ -coordinate -2, not 2. The sign inside the bracket is the opposite sign of the centre coordinate.

Finding a missing constant

If a point lies on a circle, its coordinates satisfy the circle equation. Substitute the point in to find the missing value.

Example

Using a point on the circle to find a constant

The circle has equation

x^2+y^2-6x+4y+k=0

and passes through $(2, 3)$ . Find $k$ , then find the centre and radius.

Substitute $x = 2$ and $y = 3$ into the equation.
$2^2+3^2-6(2)+4(3)+k=0$
Simplify to find $k$ .
$4 + 9 - 12 + 12 + k = 0$ $13 + k = 0$ $k = - 13$
Write the equation with $k = - 13$ .
$x^2+y^2-6x+4y-13=0$
Complete the square.
$x-3)^2-9+(y+2)^2-4-13=0$
Rearrange into standard form.
$x-3)^2+(y+2)^2=26$
Read off the centre and radius.
$=(3,−2),r=26\text{centre }=(3,-2),\quad r=\sqrt{26}$

Intersections with lines

To find where a line meets a circle, substitute the line equation into the circle equation. This usually creates a quadratic.

A line can meet a circle in two points, touch it once as a tangent, or miss it completely depending on the quadratic roots.

Two distinct roots means the line cuts the circle at two points.
One repeated root means the line is a tangent.
No real roots means the line misses the circle.

Example

Finding the points where a line meets a circle

The circle has centre $(2, 3)$ and radius 5. The line $y = x + 2$ intersects the circle at $A$ and $B$ . Find their coordinates.

The line crosses the circle at the two points found by solving the resulting quadratic.

Write the circle equation.
$x-2)^2+(y-3)^2=25$
Substitute $y = x + 2$ into the circle equation.
$x-2)^2+(x+2-3)^2=25$
Simplify and solve the quadratic.
$x-2)^2+(x-1)^2=25$ $x^2-4x+4+x^2-2x+1=25$ $2x^2-6x-20=0$ $x^2-3x-10=0$ $(x - 5) (x + 2) = 0$
Find the corresponding $y$ values using $y = x + 2$ .
$\Rightarrow y=7$ $\Rightarrow y=0$
State the intersection points.
$A(5,7),B(−2,0)A(5,7),\quad B(-2,0)$

Tip

Axis intersections

For intersections with the $y$ -axis, set $x = 0$ . For intersections with the $x$ -axis, set $y = 0$ .

Tangents to circles

Definition

Tangent

A tangent is a straight line that touches a circle at exactly one point. The point where it touches is called the point of contact.

The key fact is that the radius to the point of contact is perpendicular to the tangent.

A labelled diagram showing a radius meeting a tangent at right angles

If two non-vertical lines are perpendicular, their gradients multiply to -1. So if the radius has gradient $m$ , the tangent has gradient $−1m-\frac{1}{m}$ .

Example

Finding the equation of a tangent at a point

A circle has centre $(- 1, 2)$ and passes through $A (4, 4)$ . Find the equation of the tangent at $A$ in the form $a x + b y + c = 0$ .

The tangent at a point is perpendicular to the radius drawn to that point.

Find the gradient of the radius from the centre to $A$ .
$mradius=4−24−(−1)=25m_{\text{radius}}=\frac{4-2}{4-(-1)}=\frac{2}{5}$
Use the negative reciprocal for the tangent gradient.
$mtangent=−52m_{\text{tangent}}=-\frac{5}{2}$
Use the point-gradient form through $A (4, 4)$ .
$y−4=−52(x−4)y-4=-\frac{5}{2}(x-4)$
Rearrange into integer form.
$2 y - 8 = - 5 x + 20$ $5 x + 2 y - 28 = 0$

Common Mistake

Using the radius gradient as the tangent gradient

The tangent is not parallel to the radius. It is perpendicular to it, so you must use the negative reciprocal gradient.

Tangency using distance from a point to a line

Sometimes you are given a centre and a tangent line, but not the radius. The radius is the perpendicular distance from the centre to the tangent line.

For a line $a x + b y + c = 0$ , the distance from $x_1,y_1)$ to the line is:

∣ax1+by1+c∣a2+b2\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}

Example

Finding a circle from its centre and a tangent line

A circle has centre $(6, - 1)$ . The line $3 x + y - 4 = 0$ is tangent to the circle. Find the equation of the circle.

Use the distance from the centre to the tangent line to find the radius.
$r=∣3(6)+(−1)−4∣32+12r=\frac{|3(6)+(-1)-4|}{\sqrt{3^2+1^2}}$
Simplify.
$r=∣18−1−4∣10=1310r=\frac{|18-1-4|}{\sqrt{10}}=\frac{13}{\sqrt{10}}$
Square the radius for the circle equation.
$r2=(1310)2=16910r^2=\left(\frac{13}{\sqrt{10}}\right)^2=\frac{169}{10}$
Write the equation using centre $(6, - 1)$ .
$(x−6)2+(y+1)2=16910(x-6)^2+(y+1)^2=\frac{169}{10}$

Exam technique

In the exam

Put circle equations into standard form as early as possible, because the centre and radius then become easy to read.
For tangents, draw a quick sketch and mark the radius at 90° to the tangent.
When substituting a line into a circle, expect a quadratic and use the discriminant if the question asks about tangency or ranges.

Self review

Check yourself

Can you find the centre and radius of $x^2+y^2-8x+6y=12$ ?
If a circle has centre $(4, - 2)$ and passes through $(1, 2)$ , can you write its equation?
Can you find the tangent gradient at a point once you know the gradient of the radius?

Circles

What you'll learn

Prerequisites: coordinate-geometry tools

The standard equation of a circle

Circles from a diameter

Completing the square

Finding a missing constant

Intersections with lines

Tangents to circles

Tangency using distance from a point to a line

Recap questions

You've reached the end

Circles

What you'll learn

Prerequisites: coordinate-geometry tools

The standard equation of a circle

Circles from a diameter

Completing the square

Finding a missing constant

Intersections with lines

Tangents to circles

Tangency using distance from a point to a line

Recap questions

You've reached the end

Circles

What you'll learn

Prerequisites: coordinate-geometry tools

The standard equation of a circle

Circles from a diameter

Completing the square

Finding a missing constant

Intersections with lines

Tangents to circles

Tangency using distance from a point to a line

Recap questions

You've reached the end

Circles Revision Guide

Circles

What you'll learn

Prerequisites: coordinate-geometry tools

The standard equation of a circle

Circles from a diameter

Completing the square

Finding a missing constant

Intersections with lines

Tangents to circles

Tangency using distance from a point to a line

Recap questions

You've reached the end

Circles Revision Guide