Circles Through Three Points | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Two chord bisectors locate the centre

Join two different pairs of points to make two chords.
Find the perpendicular bisector of each chord.
Solve the two bisector equations simultaneously. Their intersection is the circle centre.
Find $r^2$ from the centre to any one of the three points, then write the circle equation.

Use two chords that are not parallel. If the three given points are collinear, their chord bisectors do not meet at one finite centre, so no circle passes through all three.

Keep the two bisectors separate

Worked example

A circle passes through $P(-1,4)$ , $Q(1,6)$ and $R(5,4)$ . Find its equation

Coursebook Worked example 3.13.

For $PQ$ , the midpoint is $(0,5)$ and the gradient is $1$ . Its perpendicular bisector is $y=-x+5$ .
For $QR$ , the midpoint is $(3,5)$ and the gradient is $-1/2$ . Its perpendicular bisector is $y=2x-1$ .
Solve $-x+5=2x-1$ . The centre is $C(2,3)$ .
$r^2=CP^2=(-1-2)^2+(4-3)^2=10$ .

Answer

(x-2)^2+(y-3)^2=10

Swipe left or right to see the whole diagram →

The perpendicular bisectors of PQ and QR meet at the circle centre C(2, 3).

An algebraic route when bisectors are awkward

Start with $(x-a)^2+(y-b)^2=r^2$ and substitute all three points. For the same $P,Q,R$ above, expanding gives:

$\begin{aligned}P:&\quad a^2+b^2+2a-8b+17=r^2\\Q:&\quad a^2+b^2-2a-12b+37=r^2\\R:&\quad a^2+b^2-10a-8b+41=r^2\end{aligned}$

Subtract whole equations: the $a^2,b^2,r^2$ terms cancel.

$Q-P:\quad -4a-4b+20=0\quad\Rightarrow\quad a+b=5$

$R-P:\quad -12a+24=0\quad\Rightarrow\quad a=2$

Hence $b=3$ . Substitute into the equation for $P$ to get $r^2=10$ . Use this route when subtraction is simpler than finding two bisectors.

Construct the circle

Your turnCoursebook Exercise 3D Q9

Find the circle through $O(0,0)$ , $A(8,4)$ and $B(6,6)$ .

Show worked answer

Answer

The perpendicular bisectors of $OA$ and $OB$ are $y=-2x+10$ and $y=-x+6$ .

They meet at $(4,2)$ . Then $r^2=4^2+2^2=20$ , so the circle is $(x-4)^2+(y-2)^2=20$ .

Two chord bisectors locate the centre

Join two different pairs of points to make two chords.
Find the perpendicular bisector of each chord.
Solve the two bisector equations simultaneously. Their intersection is the circle centre.
Find $r^2$ from the centre to any one of the three points, then write the circle equation.

Use two chords that are not parallel. If the three given points are collinear, their chord bisectors do not meet at one finite centre, so no circle passes through all three.

Keep the two bisectors separate

Worked example

A circle passes through $P(-1,4)$ , $Q(1,6)$ and $R(5,4)$ . Find its equation

Coursebook Worked example 3.13.

For $PQ$ , the midpoint is $(0,5)$ and the gradient is $1$ . Its perpendicular bisector is $y=-x+5$ .
For $QR$ , the midpoint is $(3,5)$ and the gradient is $-1/2$ . Its perpendicular bisector is $y=2x-1$ .
Solve $-x+5=2x-1$ . The centre is $C(2,3)$ .
$r^2=CP^2=(-1-2)^2+(4-3)^2=10$ .

Answer

(x-2)^2+(y-3)^2=10

Swipe left or right to see the whole diagram →

The perpendicular bisectors of PQ and QR meet at the circle centre C(2, 3).

An algebraic route when bisectors are awkward

Start with $(x-a)^2+(y-b)^2=r^2$ and substitute all three points. For the same $P,Q,R$ above, expanding gives:

$\begin{aligned}P:&\quad a^2+b^2+2a-8b+17=r^2\\Q:&\quad a^2+b^2-2a-12b+37=r^2\\R:&\quad a^2+b^2-10a-8b+41=r^2\end{aligned}$

Subtract whole equations: the $a^2,b^2,r^2$ terms cancel.

$Q-P:\quad -4a-4b+20=0\quad\Rightarrow\quad a+b=5$

$R-P:\quad -12a+24=0\quad\Rightarrow\quad a=2$

Hence $b=3$ . Substitute into the equation for $P$ to get $r^2=10$ . Use this route when subtraction is simpler than finding two bisectors.

Construct the circle

Your turnCoursebook Exercise 3D Q9

Find the circle through $O(0,0)$ , $A(8,4)$ and $B(6,6)$ .

Show worked answer

Answer

The perpendicular bisectors of $OA$ and $OB$ are $y=-2x+10$ and $y=-x+6$ .

They meet at $(4,2)$ . Then $r^2=4^2+2^2=20$ , so the circle is $(x-4)^2+(y-2)^2=20$ .