6.2 Equation of a Circle

What you'll learn

Write a circle equation in centre-radius form: the version that shows the centre and radius immediately.
Use completing the square to find the centre and radius from an expanded equation.
Find a circle equation from a centre and a point, or from a diameter.
Handle lines that meet or touch a circle, including exact intersections and tangents.

1. The basic geometry of a circle

A circle is all about distance. Every point on the circle is the same distance from one fixed point.

Definition

Circle, centre, radius and tangent

A circle is the set of all points a fixed distance from a fixed point.
The centre is the fixed point in the middle of the circle.
The radius is the fixed distance from the centre to any point on the circle.
A tangent is a straight line that touches a circle at exactly one point, called the point of contact.

The diagram shows the main geometry: every point on the circle is the same distance from the centre, and a tangent is at right angles to the radius at the point of contact.

Coordinate diagram of a circle with centre, radius and tangent labelled

Key Idea

Centre-radius form

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

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

Notice the signs: if the centre is $(- 2, 3)$ , the bracket is $x+2)^2$ , not $x-2)^2$ .

Example

Writing an equation from centre and radius

A circle has centre $(- 2, 3)$ and radius 4. Find its equation.

Start with the centre-radius form:
$x-a)^2+(y-b)^2=r^2$
Here $a = - 2$ , $b = 3$ and $r = 4$ , so substitute these values:
$x-(-2))^2+(y-3)^2=4^2$
Simplify the brackets and the radius squared:
$x+2)^2+(y-3)^2=16$

2. Finding the radius from a point on the circle

The distance formula gives the distance between two points. If the points are $x_1,y_1)$ and $x_2,y_2)$ , then

distance=(x2−x1)2+(y2−y1)2\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For circle equations, it is often quicker to find $r^2$ directly, because the equation uses $r^2$ .

Example

Centre and a point on the circle

A circle has centre $(2, - 3)$ and passes through the point $(5, 1)$ . Find its equation.

A coordinate sketch showing how the radius squared comes from the horizontal and vertical differences between the centre and a point on the circle.

Find $r^2$ using the distance from the centre to the point:
$r2=(5−2)2+(1−(−3))2=32+42=25\begin{aligned} r^2&=(5-2)^2+(1-(-3))^2\\ &=3^2+4^2\\ &=25 \end{aligned}$
Substitute the centre and $r^2=25$ into the circle equation:
$x-2)^2+(y+3)^2=25$

Tip

Use radius squared when possible

If you are writing the equation of a circle, you do not always need to find $r$ . Finding $r^2$ is enough.

3. Expanded form and completing the square

Circle equations are not always given neatly in centre-radius form. They may be expanded, such as

x^2+y^2+6x-8y=11

To recover the centre and radius, use completing the square.

Definition

Completing the square

Completing the square means rewriting a quadratic expression using a squared bracket, such as changing $x^2+6x$ into $x+3)^2-9$ .

The pattern is: halve the coefficient of $x$ , put it inside the bracket, then subtract the square you have added.

x2+6x=(x+3)2−9y2−8y=(y−4)2−16\begin{aligned} x^2+6x&=(x+3)^2-9\\ y^2-8y&=(y-4)^2-16 \end{aligned}

Example

Finding centre and radius from an expanded equation

Find the centre and radius of the circle

Completing the square converts the expanded equation into centre-radius form, revealing centre and radius.

x^2+y^2+6x-8y=11

Group the $x$ terms and the $y$ terms:
$x^2+6x)+(y^2-8y)=11$
Complete the square for each pair of terms:
$x+3)^2-9+(y-4)^2-16=11$
Move the constants to the right-hand side:
$x+3)^2+(y-4)^2=36$
Compare with $x-a)^2+(y-b)^2=r^2$ . The centre is $(- 3, 4)$ and the radius is 6.

Common Mistake

The centre has opposite signs

In $x+3)^2+(y-4)^2=36$ , the centre is $(- 3, 4)$ , not $(3, - 4)$ . The signs inside the brackets are opposite to the centre coordinates.

Common Mistake

Check it is a real circle

After completing the square, the right-hand side must be non-negative. A negative value for $r^2$ means there is no real circle.

4. Using a point to find an unknown constant

Sometimes the equation contains an unknown constant. If a point lies on the circle, its coordinates must satisfy the equation.

Example

Finding an unknown constant

The circle

A line intersects a circle where the two equations are satisfied at the same coordinates.

The given point lies on the circle, so substituting its coordinates determines the unknown constant before the centre and radius are found.

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

passes through the point $(3, 5)$ . Find $k$ , then find the centre and radius.

Substitute $x = 3$ and $y = 5$ into the equation:
$3^2+5^2+2(3)-4(5)+k=0$
Simplify to find $k$ :
$20+k=0⇒k=−2020+k=0\Rightarrow k=-20$
Substitute $k = - 20$ and complete the square:
$x2+y2+2x−4y−20=0(x+1)2−1+(y−2)2−4−20=0(x+1)2+(y−2)2=25\begin{aligned} x^2+y^2+2x-4y-20&=0\\ (x+1)^2-1+(y-2)^2-4-20&=0\\ (x+1)^2+(y-2)^2&=25 \end{aligned}$
The centre is $(- 1, 2)$ and the radius is 5.

5. Intersections with a line or an axis

If a line meets a circle, the intersection points satisfy both equations. So you substitute the line equation into the circle equation.

For the coordinate axes:

On the x-axis, $y = 0$ .
On the y-axis, $x = 0$ .

Example

Finding exact intersections with a line

The circle

x-2)^2+(y+1)^2=25

meets the line $y = 2 x - 3$ at two points. Find their exact coordinates.

Substitute $y = 2 x - 3$ into the circle equation. Since $y + 1 = 2 x - 2$ :
$x-2)^2+(2x-2)^2=25$
Expand and simplify:
$x2−4x+4+4x2−8x+4=255x2−12x−17=0\begin{aligned} x^2-4x+4+4x^2-8x+4&=25\\ 5x^2-12x-17&=0 \end{aligned}$
Factorise the quadratic:
$(5 x - 17) (x + 1) = 0$
So the possible $x$ -values are $x=175x=\frac{17}{5}$ and $x = - 1$ .
Use $y = 2 x - 3$ . The intersection points are $(175,195)\left(\frac{17}{5},\frac{19}{5}\right)$ and $(- 1, - 5)$ .

6. Tangents to a circle

The gradient of a straight line is its slope. For two non-vertical perpendicular lines, their gradients multiply to -1.

Key Idea

Radius meets tangent at 90°

The radius to the point of contact is perpendicular to the tangent. So find the radius gradient first, then use the negative reciprocal for the tangent gradient.

Example

Finding the equation of a tangent

A circle has centre $(1, 4)$ . The point $P (- 2, 0)$ lies on the circle. Find the tangent to the circle at $P$ , giving your answer in the form $a x + b y + c = 0$ .

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

Find the gradient of the radius from the centre to $P$ :
$m=0−4−2−1=−4−3=43m=\frac{0-4}{-2-1}=\frac{-4}{-3}=\frac{4}{3}$
The tangent gradient is the negative reciprocal:
$−34-\frac{3}{4}$
Use the straight-line equation through $P (- 2, 0)$ :
$y−0=−34(x+2)y-0=-\frac{3}{4}(x+2)$
Rearrange into $a x + b y + c = 0$ :
$4y=−3x−6⇒3x+4y+6=04y=-3x-6\Rightarrow 3x+4y+6=0$

Tip

Horizontal and vertical tangents

If the radius is vertical, the tangent is horizontal. If the radius is horizontal, the tangent is vertical. The negative reciprocal rule only works when both gradients are finite.

Tangent length from an external point

If a point outside a circle has a tangent to the circle, the radius to the point of contact is perpendicular to the tangent. That gives a right-angled triangle, so you can use Pythagoras’ theorem.

Example

Exact length of a tangent segment

A circle has centre $C (- 1, 2)$ and radius 4. A tangent from the point $A (7, 7)$ touches the circle at $B$ . Find the exact length $A B$ .

The radius to the point of contact and the tangent form a right-angled triangle, so Pythagoras gives the tangent length.

Find $AC^2$ :
$AC^2=(7-(-1))^2+(7-2)^2=8^2+5^2=89$
Since $C B$ is a radius, $C B = 4$ , so $CB^2=16$ .
Triangle $A C B$ is right-angled at $B$ , so use Pythagoras:
$AB^2=AC^2-CB^2=89-16=73$
Therefore the exact length is:
$AB=73AB=\sqrt{73}$

7. Diameter questions

A diameter is a line segment joining two points on the circle and passing through the centre. If you know the endpoints of a diameter, the centre is the midpoint.

Example

Equation from a diameter

The points $A (- 4, 6)$ and $B (8, - 2)$ are endpoints of a diameter of a circle. Find the equation of the circle.

The centre of a circle is the midpoint of the endpoints of a diameter.

Find the midpoint of $A$ and $B$ :
$(−4+82,6+(−2)2)=(2,2)\left(\frac{-4+8}{2},\frac{6+(-2)}{2}\right)=(2,2)$
Find $r^2$ using the centre $(2, 2)$ and one endpoint:
$r^2=(-4-2)^2+(6-2)^2=36+16=52$
Write the equation:
$x-2)^2+(y-2)^2=52$

Tip

Right angle in a semicircle

If three points lie on a circle and one angle is a right angle, the side opposite the right angle is a diameter. To show a right angle using coordinates, show the two relevant gradients multiply to -1.

An angle standing on a diameter is a right angle, which can be checked using perpendicular gradients.

Exam technique

In the exam

First put the circle into centre-radius form, or identify the centre and $r^2$ directly if it is already in that form.
For tangents, always draw or imagine the radius to the point of contact: radius and tangent are perpendicular.
Keep exact answers where possible, especially fractions and surds, unless the question asks for decimals.

Self review

Check yourself

Can you complete the square for $x^2-10x$ and explain why the centre coordinate changes sign?
If a circle has centre $(4, - 1)$ and passes through $(1, 3)$ , can you find $r^2$ without first finding $r$ ?
In a tangent question, which gradient do you find first: the tangent’s or the radius’s?

6.2 Equation of a Circle

What you'll learn

Write a circle equation in centre-radius form: the version that shows the centre and radius immediately.
Use completing the square to find the centre and radius from an expanded equation.
Find a circle equation from a centre and a point, or from a diameter.
Handle lines that meet or touch a circle, including exact intersections and tangents.

1. The basic geometry of a circle

A circle is all about distance. Every point on the circle is the same distance from one fixed point.

Definition

Circle, centre, radius and tangent

A circle is the set of all points a fixed distance from a fixed point.
The centre is the fixed point in the middle of the circle.
The radius is the fixed distance from the centre to any point on the circle.
A tangent is a straight line that touches a circle at exactly one point, called the point of contact.

The diagram shows the main geometry: every point on the circle is the same distance from the centre, and a tangent is at right angles to the radius at the point of contact.

Coordinate diagram of a circle with centre, radius and tangent labelled

Key Idea

Centre-radius form

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

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

Notice the signs: if the centre is $(- 2, 3)$ , the bracket is $x+2)^2$ , not $x-2)^2$ .

Example

Writing an equation from centre and radius

A circle has centre $(- 2, 3)$ and radius 4. Find its equation.

Start with the centre-radius form:
$x-a)^2+(y-b)^2=r^2$
Here $a = - 2$ , $b = 3$ and $r = 4$ , so substitute these values:
$x-(-2))^2+(y-3)^2=4^2$
Simplify the brackets and the radius squared:
$x+2)^2+(y-3)^2=16$

2. Finding the radius from a point on the circle

The distance formula gives the distance between two points. If the points are $x_1,y_1)$ and $x_2,y_2)$ , then

distance=(x2−x1)2+(y2−y1)2\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For circle equations, it is often quicker to find $r^2$ directly, because the equation uses $r^2$ .

Example

Centre and a point on the circle

A circle has centre $(2, - 3)$ and passes through the point $(5, 1)$ . Find its equation.

A coordinate sketch showing how the radius squared comes from the horizontal and vertical differences between the centre and a point on the circle.

Find $r^2$ using the distance from the centre to the point:
$r2=(5−2)2+(1−(−3))2=32+42=25\begin{aligned} r^2&=(5-2)^2+(1-(-3))^2\\ &=3^2+4^2\\ &=25 \end{aligned}$
Substitute the centre and $r^2=25$ into the circle equation:
$x-2)^2+(y+3)^2=25$

Tip

Use radius squared when possible

If you are writing the equation of a circle, you do not always need to find $r$ . Finding $r^2$ is enough.

3. Expanded form and completing the square

Circle equations are not always given neatly in centre-radius form. They may be expanded, such as

x^2+y^2+6x-8y=11

To recover the centre and radius, use completing the square.

Definition

Completing the square

Completing the square means rewriting a quadratic expression using a squared bracket, such as changing $x^2+6x$ into $x+3)^2-9$ .

The pattern is: halve the coefficient of $x$ , put it inside the bracket, then subtract the square you have added.

x2+6x=(x+3)2−9y2−8y=(y−4)2−16\begin{aligned} x^2+6x&=(x+3)^2-9\\ y^2-8y&=(y-4)^2-16 \end{aligned}

Example

Finding centre and radius from an expanded equation

Find the centre and radius of the circle

Completing the square converts the expanded equation into centre-radius form, revealing centre and radius.

x^2+y^2+6x-8y=11

Group the $x$ terms and the $y$ terms:
$x^2+6x)+(y^2-8y)=11$
Complete the square for each pair of terms:
$x+3)^2-9+(y-4)^2-16=11$
Move the constants to the right-hand side:
$x+3)^2+(y-4)^2=36$
Compare with $x-a)^2+(y-b)^2=r^2$ . The centre is $(- 3, 4)$ and the radius is 6.

Common Mistake

The centre has opposite signs

In $x+3)^2+(y-4)^2=36$ , the centre is $(- 3, 4)$ , not $(3, - 4)$ . The signs inside the brackets are opposite to the centre coordinates.

Common Mistake

Check it is a real circle

After completing the square, the right-hand side must be non-negative. A negative value for $r^2$ means there is no real circle.

4. Using a point to find an unknown constant

Sometimes the equation contains an unknown constant. If a point lies on the circle, its coordinates must satisfy the equation.

Example

Finding an unknown constant

The circle

A line intersects a circle where the two equations are satisfied at the same coordinates.

The given point lies on the circle, so substituting its coordinates determines the unknown constant before the centre and radius are found.

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

passes through the point $(3, 5)$ . Find $k$ , then find the centre and radius.

Substitute $x = 3$ and $y = 5$ into the equation:
$3^2+5^2+2(3)-4(5)+k=0$
Simplify to find $k$ :
$20+k=0⇒k=−2020+k=0\Rightarrow k=-20$
Substitute $k = - 20$ and complete the square:
$x2+y2+2x−4y−20=0(x+1)2−1+(y−2)2−4−20=0(x+1)2+(y−2)2=25\begin{aligned} x^2+y^2+2x-4y-20&=0\\ (x+1)^2-1+(y-2)^2-4-20&=0\\ (x+1)^2+(y-2)^2&=25 \end{aligned}$
The centre is $(- 1, 2)$ and the radius is 5.

5. Intersections with a line or an axis

If a line meets a circle, the intersection points satisfy both equations. So you substitute the line equation into the circle equation.

For the coordinate axes:

On the x-axis, $y = 0$ .
On the y-axis, $x = 0$ .

Example

Finding exact intersections with a line

The circle

x-2)^2+(y+1)^2=25

meets the line $y = 2 x - 3$ at two points. Find their exact coordinates.

Substitute $y = 2 x - 3$ into the circle equation. Since $y + 1 = 2 x - 2$ :
$x-2)^2+(2x-2)^2=25$
Expand and simplify:
$x2−4x+4+4x2−8x+4=255x2−12x−17=0\begin{aligned} x^2-4x+4+4x^2-8x+4&=25\\ 5x^2-12x-17&=0 \end{aligned}$
Factorise the quadratic:
$(5 x - 17) (x + 1) = 0$
So the possible $x$ -values are $x=175x=\frac{17}{5}$ and $x = - 1$ .
Use $y = 2 x - 3$ . The intersection points are $(175,195)\left(\frac{17}{5},\frac{19}{5}\right)$ and $(- 1, - 5)$ .

6. Tangents to a circle

The gradient of a straight line is its slope. For two non-vertical perpendicular lines, their gradients multiply to -1.

Key Idea

Radius meets tangent at 90°

The radius to the point of contact is perpendicular to the tangent. So find the radius gradient first, then use the negative reciprocal for the tangent gradient.

Example

Finding the equation of a tangent

A circle has centre $(1, 4)$ . The point $P (- 2, 0)$ lies on the circle. Find the tangent to the circle at $P$ , giving your answer in the form $a x + b y + c = 0$ .

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

Find the gradient of the radius from the centre to $P$ :
$m=0−4−2−1=−4−3=43m=\frac{0-4}{-2-1}=\frac{-4}{-3}=\frac{4}{3}$
The tangent gradient is the negative reciprocal:
$−34-\frac{3}{4}$
Use the straight-line equation through $P (- 2, 0)$ :
$y−0=−34(x+2)y-0=-\frac{3}{4}(x+2)$
Rearrange into $a x + b y + c = 0$ :
$4y=−3x−6⇒3x+4y+6=04y=-3x-6\Rightarrow 3x+4y+6=0$

Tip

Horizontal and vertical tangents

If the radius is vertical, the tangent is horizontal. If the radius is horizontal, the tangent is vertical. The negative reciprocal rule only works when both gradients are finite.

Tangent length from an external point

If a point outside a circle has a tangent to the circle, the radius to the point of contact is perpendicular to the tangent. That gives a right-angled triangle, so you can use Pythagoras’ theorem.

Example

Exact length of a tangent segment

A circle has centre $C (- 1, 2)$ and radius 4. A tangent from the point $A (7, 7)$ touches the circle at $B$ . Find the exact length $A B$ .

The radius to the point of contact and the tangent form a right-angled triangle, so Pythagoras gives the tangent length.

Find $AC^2$ :
$AC^2=(7-(-1))^2+(7-2)^2=8^2+5^2=89$
Since $C B$ is a radius, $C B = 4$ , so $CB^2=16$ .
Triangle $A C B$ is right-angled at $B$ , so use Pythagoras:
$AB^2=AC^2-CB^2=89-16=73$
Therefore the exact length is:
$AB=73AB=\sqrt{73}$

7. Diameter questions

A diameter is a line segment joining two points on the circle and passing through the centre. If you know the endpoints of a diameter, the centre is the midpoint.

Example

Equation from a diameter

The points $A (- 4, 6)$ and $B (8, - 2)$ are endpoints of a diameter of a circle. Find the equation of the circle.

The centre of a circle is the midpoint of the endpoints of a diameter.

Find the midpoint of $A$ and $B$ :
$(−4+82,6+(−2)2)=(2,2)\left(\frac{-4+8}{2},\frac{6+(-2)}{2}\right)=(2,2)$
Find $r^2$ using the centre $(2, 2)$ and one endpoint:
$r^2=(-4-2)^2+(6-2)^2=36+16=52$
Write the equation:
$x-2)^2+(y-2)^2=52$

Tip

Right angle in a semicircle

An angle standing on a diameter is a right angle, which can be checked using perpendicular gradients.

Exam technique

In the exam

First put the circle into centre-radius form, or identify the centre and $r^2$ directly if it is already in that form.
For tangents, always draw or imagine the radius to the point of contact: radius and tangent are perpendicular.
Keep exact answers where possible, especially fractions and surds, unless the question asks for decimals.

Self review

Check yourself

Can you complete the square for $x^2-10x$ and explain why the centre coordinate changes sign?
If a circle has centre $(4, - 1)$ and passes through $(1, 3)$ , can you find $r^2$ without first finding $r$ ?
In a tangent question, which gradient do you find first: the tangent’s or the radius’s?

6.2 Equation of a Circle

What you'll learn

1. The basic geometry of a circle

2. Finding the radius from a point on the circle

3. Expanded form and completing the square

4. Using a point to find an unknown constant

5. Intersections with a line or an axis

6. Tangents to a circle

Tangent length from an external point

7. Diameter questions

Recap questions

6.2 Equation of a Circle

What you'll learn

1. The basic geometry of a circle

2. Finding the radius from a point on the circle

3. Expanded form and completing the square

4. Using a point to find an unknown constant

5. Intersections with a line or an axis

6. Tangents to a circle

Tangent length from an external point

7. Diameter questions

Recap questions

6.2 Equation of a Circle

What you'll learn

1. The basic geometry of a circle

2. Finding the radius from a point on the circle

3. Expanded form and completing the square

4. Using a point to find an unknown constant

5. Intersections with a line or an axis

6. Tangents to a circle

Tangent length from an external point

7. Diameter questions

Recap questions

6.2 Equation of a Circle Revision Guide

6.2 Equation of a Circle

What you'll learn

1. The basic geometry of a circle

2. Finding the radius from a point on the circle

3. Expanded form and completing the square

4. Using a point to find an unknown constant

5. Intersections with a line or an axis

6. Tangents to a circle

Tangent length from an external point

7. Diameter questions

Recap questions

6.2 Equation of a Circle Revision Guide