The Equation of a Circle | CAIE 9709 Pure Mathematics 1 Notes

Chapters

Pure Mathematics 1

Mathematics9709

The circle equation is a distance equation

Let the centre be $C(a,b)$ and a point on the circle be $P(x,y)$ . Since $CP=r$ , the squared distance formula gives

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

Swipe left or right to see the whole diagram →

Every point P(x, y) on the circle is exactly r units from the fixed centre C(a, b).

The centre signs are opposite to the signs inside the brackets. For $(x+3)^2+(y-2)^2=49$ , the centre is $(-3,2)$ .
The right side is $r^2$ , not $r$ . Here the radius is $\sqrt{49}=7$ .

Build a circle from the given information

The target form needs a centre and $r^2$ . If the centre and one point are given, calculate their squared distance. If the ends of a are given, the centre is their midpoint: a diameter passes through the centre and is two radii long.

Worked example

$A(-1,-2)$ and $B(7,4)$ are opposite ends of a diameter. Find the circle

(9709/11 M/J 2020 Q10(a))

Centre = midpoint: $C=\left(\frac{-1+7}{2},\frac{-2+4}{2}\right)=(3,1)$ .
Keep the radius squared: $r^2=(7-3)^2+(4-1)^2=16+9=25$ .
Substitute the centre and $r^2$ into the standard form.

Answer

(x-3)^2+(y-1)^2=25

Swipe left or right to see the whole diagram →

AB is a diameter, so its midpoint C is the centre and CA = CB = 5.

Fast checks before you finish

Substitute a given point. It must satisfy the circle equation.
A circle centred at $(a,b)$ has extreme x-values $a-r$ and $a+r$ , and extreme y-values $b-r$ and $b+r$ .
A sketch should show the centre, radius information and relevant intercepts or contact points. It is not evidence for an exact coordinate.

Use centre and radius squared

Your turnIndependent check

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

Show worked answer

Answer

$r^2=(5-2)^2+(-1-(-5))^2=3^2+4^2=25$

$(x-2)^2+(y+5)^2=25$

The circle equation is a distance equation

Let the centre be $C(a,b)$ and a point on the circle be $P(x,y)$ . Since $CP=r$ , the squared distance formula gives

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

Swipe left or right to see the whole diagram →

Every point P(x, y) on the circle is exactly r units from the fixed centre C(a, b).

The centre signs are opposite to the signs inside the brackets. For $(x+3)^2+(y-2)^2=49$ , the centre is $(-3,2)$ .
The right side is $r^2$ , not $r$ . Here the radius is $\sqrt{49}=7$ .

Build a circle from the given information

Worked example

$A(-1,-2)$ and $B(7,4)$ are opposite ends of a diameter. Find the circle

(9709/11 M/J 2020 Q10(a))

Centre = midpoint: $C=\left(\frac{-1+7}{2},\frac{-2+4}{2}\right)=(3,1)$ .
Keep the radius squared: $r^2=(7-3)^2+(4-1)^2=16+9=25$ .
Substitute the centre and $r^2$ into the standard form.

Answer

(x-3)^2+(y-1)^2=25

Swipe left or right to see the whole diagram →

AB is a diameter, so its midpoint C is the centre and CA = CB = 5.

Fast checks before you finish

Substitute a given point. It must satisfy the circle equation.
A circle centred at $(a,b)$ has extreme x-values $a-r$ and $a+r$ , and extreme y-values $b-r$ and $b+r$ .
A sketch should show the centre, radius information and relevant intercepts or contact points. It is not evidence for an exact coordinate.

Use centre and radius squared

Your turnIndependent check

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

Show worked answer

Answer

$r^2=(5-2)^2+(-1-(-5))^2=3^2+4^2=25$

$(x-2)^2+(y+5)^2=25$