1.3.5 Centre and radius by completing the square

What you'll learn

How the equation of a circle encodes its centre and radius.
How to complete the square separately in $x$ and $y$ .
How to rewrite an expanded circle equation in standard form.
How to recognise equations that do not represent a real circle.

The standard equation of a circle

A circle is the set of all points that are a fixed distance from one particular point. That 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

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

This is called the standard form of the equation of a circle.

A circle with centre C at (a, b), radius r and a right-angled triangle showing horizontal displacement x-a and vertical displacement y-b

The equation follows from Pythagoras' theorem. For any point $(x, y)$ on the circle:

the horizontal displacement from the centre is $x - a$ ;
the vertical displacement from the centre is $y - b$ ;
the distance from the centre is $r$ .

Therefore,

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

Definition

Reading standard form

From $x-a)^2+(y-b)^2=r^2$ , the centre is $(a, b)$ and the radius is the positive value $r$ .

Example

Reading the centre and radius

Find the centre and radius of

x-4)^2+(y+3)^2=49.

Compare the equation with $x-a)^2+(y-b)^2=r^2$ .
Since $x - 4 = x - a$ , we have $a = 4$ . Write $y + 3$ as $y - (- 3)$ , so $b = - 3$ .
Since $r^2=49$ and a radius is positive, $r = 7$ . The centre is $(4, - 3)$ and the radius is 7.

Common Mistake

The signs of the centre

The signs inside the brackets are opposite to the coordinates of the centre. For example, $x+5)^2$ gives an $x$ -coordinate of $- 5$ , not 5.

The prerequisite: completing the square

An expression such as $x^2+6x$ is not yet a perfect square. To complete the square means rewriting a quadratic expression using a squared bracket.

The key identity is

x2+px=(x+p2)2−(p2)2.x^2+px=\left(x+\frac{p}{2}\right)^2-\left(\frac{p}{2}\right)^2.

You halve the coefficient of $x$ to form the bracket, then subtract the square of that number to keep the expression unchanged.

For example,

x^2+6x=(x+3)^2-9,

because expanding $x+3)^2$ produces $x^2+6x+9$ , so 9 must be subtracted again.

Key Idea

Half, then square

To complete the square for $x^2+px$ , halve $p$ for the bracket and subtract the square of the number you inserted.

Example

Completing two quadratic expressions

Complete the square for $x^2-8x+y^2+10y$ .

For $x^2-8x$ , halve $- 8$ to get $- 4$ :
$x^2-8x=(x-4)^2-16.$
For $y^2+10y$ , halve 10 to get 5:
$y^2+10y=(y+5)^2-25.$
Combining the results gives
$x^2-8x+y^2+10y=(x-4)^2+(y+5)^2-41.$

Finding the centre and radius

A circle may initially be given in the expanded form

x^2+y^2+Dx+Ey+F=0.

To identify its centre and radius, complete the square in $x$ and $y$ separately, then rearrange into standard form.

A reliable method

Group the $x$ -terms and the $y$ -terms.
Move the constant term to the other side.
Complete the square in $x$ .
Complete the square in $y$ .
Rearrange into $x-a)^2+(y-b)^2=r^2$ .
Read off the centre and take the positive square root to find the radius.

Example

Finding a centre and radius

Find the centre and radius of the circle

x^2+y^2-6x+4y-12=0.

Group the variable terms and move the constant:
$x^2-6x)+(y^2+4y)=12.$
Complete the square in $x$ :
$x^2-6x=(x-3)^2-9.$
Complete the square in $y$ :
$y^2+4y=(y+2)^2-4.$
Substitute these forms into the equation:
$x-3)^2-9+(y+2)^2-4=12.$
Add 13 to both sides:
$x-3)^2+(y+2)^2=25.$
Comparing with standard form, the centre is $(3, - 2)$ and $r^2=25$ , so the radius is 5.

Tip

Check by expanding

After completing the square, expand your final equation mentally or on paper. The linear terms and constant should reproduce the original equation.

When the squared terms have a coefficient

The standard circle equation has coefficients of 1 for both $x^2$ and $y^2$ . If both squared terms have the same non-zero coefficient, divide the entire equation by that coefficient before completing the square.

Example

Removing a common coefficient

Find the centre and radius of

2x^2+2y^2+8x-12y-10=0.

Divide every term by 2:
$x^2+y^2+4x-6y-5=0.$
Move the constant and complete both squares:
$(x2+4x)+(y2−6y)=5,(x+2)2−4+(y−3)2−9=5.\begin{aligned} (x^2+4x)+(y^2-6y)&=5,\\ (x+2)^2-4+(y-3)^2-9&=5. \end{aligned}$
Move the correction terms to the right:
$x+2)^2+(y-3)^2=18.$
The centre is $(- 2, 3)$ and
$r=18=32.r=\sqrt{18}=3\sqrt{2}.$

Common Mistake

Checking that it is a circle

For a circle, the coefficients of $x^2$ and $y^2$ must be equal after simplification, and there must be no $x y$ term. Unequal coefficients usually describe a different curve.

A direct formula

Completing the square in the general equation

x^2+y^2+Dx+Ey+F=0

gives

(x+D2)2+(y+E2)2=D24+E24−F.\left(x+\frac{D}{2}\right)^2+ \left(y+\frac{E}{2}\right)^2 = \frac{D^2}{4}+\frac{E^2}{4}-F.

Therefore, the centre is

(−D2,−E2),\left(-\frac{D}{2},-\frac{E}{2}\right),

and the radius is

r=D24+E24−F.r=\sqrt{\frac{D^2}{4}+\frac{E^2}{4}-F}.

This formula is useful for checking your answer, but completing the square shows your reasoning clearly and is less likely to produce sign errors.

Does the equation represent a real circle?

Once the equation is in the form

x-a)^2+(y-b)^2=R,

the value of $R$ determines what the equation represents:

If $R > 0$ , it is a circle with radius $R\sqrt{R}$ .
If $R = 0$ , the equation represents only the single point $(a, b)$ .
If $R < 0$ , there are no real points satisfying the equation, because a sum of two squares cannot be negative.

Example

Recognising that no real circle exists

Determine what is represented by

x^2+y^2-4x+6y+20=0.

Complete both squares:
$x-2)^2-4+(y+3)^2-9+20=0.$
Collect the constants and rearrange:
$x-2)^2+(y+3)^2=-7.$
The left-hand side is a sum of squares, so it cannot be negative for real $x$ and $y$ . Therefore, the equation has no real points and does not represent a real circle.

Exam technique

In the exam

Make the coefficients of $x^2$ and $y^2$ equal to 1 before completing the square.
Complete the square separately for the $x$ -terms and $y$ -terms, showing the correction constants.
Check the signs when reading the centre, and take the positive square root when finding the radius.
Inspect the value of $r^2$ : a negative value means there is no real circle.

Self review

Check yourself

What are the centre and radius of $x+2)^2+(y-7)^2=36$ ?
Can you rewrite $x^2+y^2+10x-2y-10=0$ in standard circle form?
What does it mean if completing the square produces $x-a)^2+(y-b)^2=-4$ ?

What you'll learn

How the equation of a circle encodes its centre and radius.
How to complete the square separately in $x$ and $y$ .
How to rewrite an expanded circle equation in standard form.
How to recognise equations that do not represent a real circle.

The standard equation of a circle

A circle is the set of all points that are a fixed distance from one particular point. That 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

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

This is called the standard form of the equation of a circle.

A circle with centre C at (a, b), radius r and a right-angled triangle showing horizontal displacement x-a and vertical displacement y-b

The equation follows from Pythagoras' theorem. For any point $(x, y)$ on the circle:

the horizontal displacement from the centre is $x - a$ ;
the vertical displacement from the centre is $y - b$ ;
the distance from the centre is $r$ .

Therefore,

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

Definition

Reading standard form

From $x-a)^2+(y-b)^2=r^2$ , the centre is $(a, b)$ and the radius is the positive value $r$ .

Example

Reading the centre and radius

Find the centre and radius of

x-4)^2+(y+3)^2=49.

Compare the equation with $x-a)^2+(y-b)^2=r^2$ .
Since $x - 4 = x - a$ , we have $a = 4$ . Write $y + 3$ as $y - (- 3)$ , so $b = - 3$ .
Since $r^2=49$ and a radius is positive, $r = 7$ . The centre is $(4, - 3)$ and the radius is 7.

Common Mistake

The signs of the centre

The signs inside the brackets are opposite to the coordinates of the centre. For example, $x+5)^2$ gives an $x$ -coordinate of $- 5$ , not 5.

The prerequisite: completing the square

An expression such as $x^2+6x$ is not yet a perfect square. To complete the square means rewriting a quadratic expression using a squared bracket.

The key identity is

x2+px=(x+p2)2−(p2)2.x^2+px=\left(x+\frac{p}{2}\right)^2-\left(\frac{p}{2}\right)^2.

You halve the coefficient of $x$ to form the bracket, then subtract the square of that number to keep the expression unchanged.

For example,

x^2+6x=(x+3)^2-9,

because expanding $x+3)^2$ produces $x^2+6x+9$ , so 9 must be subtracted again.

Key Idea

Half, then square

To complete the square for $x^2+px$ , halve $p$ for the bracket and subtract the square of the number you inserted.

Example

Completing two quadratic expressions

Complete the square for $x^2-8x+y^2+10y$ .

For $x^2-8x$ , halve $- 8$ to get $- 4$ :
$x^2-8x=(x-4)^2-16.$
For $y^2+10y$ , halve 10 to get 5:
$y^2+10y=(y+5)^2-25.$
Combining the results gives
$x^2-8x+y^2+10y=(x-4)^2+(y+5)^2-41.$

Finding the centre and radius

A circle may initially be given in the expanded form

x^2+y^2+Dx+Ey+F=0.

To identify its centre and radius, complete the square in $x$ and $y$ separately, then rearrange into standard form.

A reliable method

Group the $x$ -terms and the $y$ -terms.
Move the constant term to the other side.
Complete the square in $x$ .
Complete the square in $y$ .
Rearrange into $x-a)^2+(y-b)^2=r^2$ .
Read off the centre and take the positive square root to find the radius.

Example

Finding a centre and radius

Find the centre and radius of the circle

x^2+y^2-6x+4y-12=0.

Group the variable terms and move the constant:
$x^2-6x)+(y^2+4y)=12.$
Complete the square in $x$ :
$x^2-6x=(x-3)^2-9.$
Complete the square in $y$ :
$y^2+4y=(y+2)^2-4.$
Substitute these forms into the equation:
$x-3)^2-9+(y+2)^2-4=12.$
Add 13 to both sides:
$x-3)^2+(y+2)^2=25.$
Comparing with standard form, the centre is $(3, - 2)$ and $r^2=25$ , so the radius is 5.

Tip

Check by expanding

After completing the square, expand your final equation mentally or on paper. The linear terms and constant should reproduce the original equation.

When the squared terms have a coefficient

Example

Removing a common coefficient

Find the centre and radius of

2x^2+2y^2+8x-12y-10=0.

Divide every term by 2:
$x^2+y^2+4x-6y-5=0.$
Move the constant and complete both squares:
$(x2+4x)+(y2−6y)=5,(x+2)2−4+(y−3)2−9=5.\begin{aligned} (x^2+4x)+(y^2-6y)&=5,\\ (x+2)^2-4+(y-3)^2-9&=5. \end{aligned}$
Move the correction terms to the right:
$x+2)^2+(y-3)^2=18.$
The centre is $(- 2, 3)$ and
$r=18=32.r=\sqrt{18}=3\sqrt{2}.$

Common Mistake

Checking that it is a circle

For a circle, the coefficients of $x^2$ and $y^2$ must be equal after simplification, and there must be no $x y$ term. Unequal coefficients usually describe a different curve.

A direct formula

Completing the square in the general equation

x^2+y^2+Dx+Ey+F=0

gives

(x+D2)2+(y+E2)2=D24+E24−F.\left(x+\frac{D}{2}\right)^2+ \left(y+\frac{E}{2}\right)^2 = \frac{D^2}{4}+\frac{E^2}{4}-F.

Therefore, the centre is

(−D2,−E2),\left(-\frac{D}{2},-\frac{E}{2}\right),

and the radius is

r=D24+E24−F.r=\sqrt{\frac{D^2}{4}+\frac{E^2}{4}-F}.

This formula is useful for checking your answer, but completing the square shows your reasoning clearly and is less likely to produce sign errors.

Does the equation represent a real circle?

Once the equation is in the form

x-a)^2+(y-b)^2=R,

the value of $R$ determines what the equation represents:

If $R > 0$ , it is a circle with radius $R\sqrt{R}$ .
If $R = 0$ , the equation represents only the single point $(a, b)$ .
If $R < 0$ , there are no real points satisfying the equation, because a sum of two squares cannot be negative.

Example

Recognising that no real circle exists

Determine what is represented by

x^2+y^2-4x+6y+20=0.

Complete both squares:
$x-2)^2-4+(y+3)^2-9+20=0.$
Collect the constants and rearrange:
$x-2)^2+(y+3)^2=-7.$
The left-hand side is a sum of squares, so it cannot be negative for real $x$ and $y$ . Therefore, the equation has no real points and does not represent a real circle.

Exam technique

In the exam

Make the coefficients of $x^2$ and $y^2$ equal to 1 before completing the square.
Complete the square separately for the $x$ -terms and $y$ -terms, showing the correction constants.
Check the signs when reading the centre, and take the positive square root when finding the radius.
Inspect the value of $r^2$ : a negative value means there is no real circle.

Self review

Check yourself

What are the centre and radius of $x+2)^2+(y-7)^2=36$ ?
Can you rewrite $x^2+y^2+10x-2y-10=0$ in standard circle form?
What does it mean if completing the square produces $x-a)^2+(y-b)^2=-4$ ?

What you'll learn

The standard equation of a circle

The prerequisite: completing the square

Finding the centre and radius

A reliable method

When the squared terms have a coefficient

A direct formula

Does the equation represent a real circle?

Review 1.3.5 Centre and radius by completing the square by teaching Genie

1.3.5 Centre and radius by completing the square

What you'll learn

The standard equation of a circle

The prerequisite: completing the square

Finding the centre and radius

A reliable method

When the squared terms have a coefficient

A direct formula

Does the equation represent a real circle?

Review 1.3.5 Centre and radius by completing the square by teaching Genie

1.3.5 Centre and radius by completing the square

What you'll learn

The standard equation of a circle

The prerequisite: completing the square

Finding the centre and radius

A reliable method

When the squared terms have a coefficient

A direct formula

Does the equation represent a real circle?

Review 1.3.5 Centre and radius by completing the square by teaching Genie

1.3.5 Centre and radius by completing the square Revision Guide

1.3.5 Centre and radius by completing the square

What you'll learn

The standard equation of a circle

The prerequisite: completing the square

Finding the centre and radius

A reliable method

When the squared terms have a coefficient

A direct formula

Does the equation represent a real circle?

Review 1.3.5 Centre and radius by completing the square by teaching Genie

1.3.5 Centre and radius by completing the square Revision Guide