Mathematics 9709 · Pure Mathematics 1

The equation of a circle

A circle is all the points that are the same fixed distance (the radius) from one fixed point (the centre).
Write that down with the distance formula and you get the circle equation. So a circle is also just Pythagoras.

Centre–radius form: $(x-a)^2+(y-b)^2=r^2$ . The centre is $(a,b)$ , but change the signs from the brackets. The radius is the square root of the right-hand side.

If you are given the multiplied-out form $x^2+y^2+2gx+2fy+c=0$ , complete the square on the $x$ and $y$ terms to find the centre and radius. Do not learn a centre formula. The textbook itself says to work it out.

Centre–radius form (it is just the distance formula)

For a point $(x,y)$ at distance $r$ from the centre $(a,b)$ , Pythagoras gives the whole equation:

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

The centre is $(a,b)$ with the signs changed, and the radius is the square root of the right side.
So $(x-3)^2+(y+1)^2=25$ has centre $(3,-1)$ and radius $5$ .

Every point on the circle is the same distance r from the centre (a, b).

The expanded form: complete the square to undo it

Exams often give you the multiplied-out form, $x^2+y^2+2gx+2fy+c=0$ (the $x^2$ and $y^2$ have the same number in front, and there is no $xy$ term), so complete the square on the $x$ - and $y$ -terms to get back to centre–radius form.

Worked example

Find the centre and radius of $x^2 + y^2 + 10x - 8y - 40 = 0$

Cambridge Pure Mathematics 1, Worked example 3.12.

Group, then complete the square for each variable: $(x+5)^2-25+(y-4)^2-16-40=0$ .
Collect the constants on the right: $(x+5)^2+(y-4)^2=81$ .
Read off, reversing the signs: $\text{centre }(-5,4)$ , $r=\sqrt{81}=9$ .

Answer

\text{centre }(-5,4),\quad r=9

A circle from a diameter

If you are given the two ends of a diameter, the centre is their midpoint. The radius is the distance from the centre to either end (half the diameter).
This is the length and midpoint topic used inside the circle.

Worked example

$A(3,\ 0)$ and $B(7,\ -4)$ are the ends of a diameter. Find the circle

Cambridge Pure Mathematics 1, Worked example 3.11.

Centre = midpoint of $AB$ : $C=\left(\tfrac{3+7}{2},\tfrac{0-4}{2}\right)=(5,-2)$ .
Radius $=CA$ , kept as $r^2$ : $r^2=(5-3)^2+(-2-0)^2=8$ .
Centre–radius form: $(x-5)^2+(y+2)^2=8$ .

Answer

(x-5)^2+(y+2)^2=8

Three right-angle facts that help with circle questions

The geometry the exam uses again and again

Cambridge uses the same three circle facts in question after question. If you know them, long algebra often becomes one line:

The tangent is perpendicular to the radius at the point where they touch. So the normal to a circle is the radius line. Its gradient is the negative reciprocal of the gradient of the tangent.
The line from the centre at a right angle to a chord cuts the chord in half. So the perpendicular bisector of any chord goes through the centre.
The angle in a semicircle is a right angle. So if $AC$ is a diameter and $B$ is on the circle, then $\angle ABC=90^\circ$ .

Each one is just the perpendicular-gradient rule from the second topic, used again inside a circle.

Where the marks go

Common mistake

The radius is

r

, not

r^2

. If the right-hand side is

25

, the radius is

5

. (9709/12 Jun 2020 Q11(a))

Common mistake

The centre signs are the opposite of the brackets.

(x+5)^2

gives

x=-5

, not

+5

Common mistake

When you complete the square, subtract the squared half twice, once for each variable, before you collect the constants. Miss one and the radius comes out wrong.

Now you try

The equation of a circle with centre $C$ is $x^2+y^2-8x+4y-5=0$ . Find the radius of the circle and the coordinates of $C$ . (9709/12 Jun 2020 Q11(a))

Your turn— tap to reveal the worked answer (9709/12 Jun 2020 Q11)

Complete the square for each variable:

$(x-4)^2-16+(y+2)^2-4-5=0$

$(x-4)^2+(y+2)^2=25$

Answer

C=(4,-2),\quad r=5

PreviousEquations of straight lines NextProblems involving intersections of lines and circles

The equation of a circle

A circle is all the points that are the same fixed distance (the radius) from one fixed point (the centre).
Write that down with the distance formula and you get the circle equation. So a circle is also just Pythagoras.

Centre–radius form: $(x-a)^2+(y-b)^2=r^2$ . The centre is $(a,b)$ , but change the signs from the brackets. The radius is the square root of the right-hand side.

Centre–radius form (it is just the distance formula)

For a point $(x,y)$ at distance $r$ from the centre $(a,b)$ , Pythagoras gives the whole equation:

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

The centre is $(a,b)$ with the signs changed, and the radius is the square root of the right side.
So $(x-3)^2+(y+1)^2=25$ has centre $(3,-1)$ and radius $5$ .

Every point on the circle is the same distance r from the centre (a, b).

The expanded form: complete the square to undo it

Worked example

Find the centre and radius of $x^2 + y^2 + 10x - 8y - 40 = 0$

Cambridge Pure Mathematics 1, Worked example 3.12.

Group, then complete the square for each variable: $(x+5)^2-25+(y-4)^2-16-40=0$ .
Collect the constants on the right: $(x+5)^2+(y-4)^2=81$ .
Read off, reversing the signs: $\text{centre }(-5,4)$ , $r=\sqrt{81}=9$ .

Answer

\text{centre }(-5,4),\quad r=9

A circle from a diameter

If you are given the two ends of a diameter, the centre is their midpoint. The radius is the distance from the centre to either end (half the diameter).
This is the length and midpoint topic used inside the circle.

Worked example

$A(3,\ 0)$ and $B(7,\ -4)$ are the ends of a diameter. Find the circle

Cambridge Pure Mathematics 1, Worked example 3.11.

Centre = midpoint of $AB$ : $C=\left(\tfrac{3+7}{2},\tfrac{0-4}{2}\right)=(5,-2)$ .
Radius $=CA$ , kept as $r^2$ : $r^2=(5-3)^2+(-2-0)^2=8$ .
Centre–radius form: $(x-5)^2+(y+2)^2=8$ .

Answer

(x-5)^2+(y+2)^2=8

Three right-angle facts that help with circle questions

The geometry the exam uses again and again

Cambridge uses the same three circle facts in question after question. If you know them, long algebra often becomes one line:

The tangent is perpendicular to the radius at the point where they touch. So the normal to a circle is the radius line. Its gradient is the negative reciprocal of the gradient of the tangent.
The line from the centre at a right angle to a chord cuts the chord in half. So the perpendicular bisector of any chord goes through the centre.
The angle in a semicircle is a right angle. So if $AC$ is a diameter and $B$ is on the circle, then $\angle ABC=90^\circ$ .

Each one is just the perpendicular-gradient rule from the second topic, used again inside a circle.

Where the marks go

Common mistake

The radius is

r

, not

r^2

. If the right-hand side is

25

, the radius is

5

. (9709/12 Jun 2020 Q11(a))

Common mistake

The centre signs are the opposite of the brackets.

(x+5)^2

gives

x=-5

, not

+5

Common mistake

When you complete the square, subtract the squared half twice, once for each variable, before you collect the constants. Miss one and the radius comes out wrong.

Now you try

The equation of a circle with centre $C$ is $x^2+y^2-8x+4y-5=0$ . Find the radius of the circle and the coordinates of $C$ . (9709/12 Jun 2020 Q11(a))

Your turn— tap to reveal the worked answer (9709/12 Jun 2020 Q11)

Complete the square for each variable:

$(x-4)^2-16+(y+2)^2-4-5=0$

$(x-4)^2+(y+2)^2=25$

Answer

C=(4,-2),\quad r=5