Pure Mathematics 1

Mathematics9709

Pure Mathematics 1

Mathematics9709

Expanded Circle Equations

A circle equation may be expanded so that its centre and radius are no longer visible. Completing the square rebuilds the centre-radius form.

Recognise the circle structure

At this level, an expanded circle has equal coefficients of $x^2$ and $y^2$ and no $xy$ term. If the common coefficient is not 1, divide the whole equation by it before completing squares.

Do not memorise a separate centre formula. Group the x-terms and y-terms, then complete each square. This exposes both the signs and the radius safely.

Complete two squares, then read the result

Worked example

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

Coursebook Worked example 3.12.

Group the variables: $(x^2+10x)+(y^2-8y)-40=0$ .
Halve 10 and -8 to get 5 and -4. Subtract their squares to keep the expression unchanged: $(x+5)^2-25+(y-4)^2-16-40=0$ .
Move the constants: $(x+5)^2+(y-4)^2=81$ .
Read the centre with opposite bracket signs and take the square root of 81 for the radius.

Answer

Centre: $(-5,4)$

Radius: $9$

Swipe left or right to see the whole diagram →

Completing the square reveals centre (-5, 4) and radius 9.

The sign of radius squared matters

$r^2>0$ : a real circle.
$r^2=0$ : one point at the centre.
$r^2<0$ : no real points satisfy the equation.

Common mistake

(x+5)^2+(y-4)^2=81

, the centre is

(-5,4)

, not

(5,-4)

, and the radius is 9, not 81.

Keep a parameter symbolic

Your turn9709/12 Oct/Nov 2024 Q8(a)

Express $x^2+y^2+px+2y+q=0$ in centre-radius form. State the centre and $r^2$ .

Show worked answer

Answer

$\left(x+\frac p2\right)^2-\frac{p^2}{4}+(y+1)^2-1+q=0$

$\left(x+\frac p2\right)^2+(y+1)^2=\frac{p^2}{4}-q+1$

Centre $\left(-\frac p2,-1\right)$ , $r^2=\frac{p^2}{4}-q+1$ .

PreviousThe Equation of a Circle NextCircles Through Three Points

Expanded Circle Equations

A circle equation may be expanded so that its centre and radius are no longer visible. Completing the square rebuilds the centre-radius form.

Recognise the circle structure

At this level, an expanded circle has equal coefficients of $x^2$ and $y^2$ and no $xy$ term. If the common coefficient is not 1, divide the whole equation by it before completing squares.

Do not memorise a separate centre formula. Group the x-terms and y-terms, then complete each square. This exposes both the signs and the radius safely.

Complete two squares, then read the result

Worked example

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

Coursebook Worked example 3.12.

Group the variables: $(x^2+10x)+(y^2-8y)-40=0$ .
Halve 10 and -8 to get 5 and -4. Subtract their squares to keep the expression unchanged: $(x+5)^2-25+(y-4)^2-16-40=0$ .
Move the constants: $(x+5)^2+(y-4)^2=81$ .
Read the centre with opposite bracket signs and take the square root of 81 for the radius.

Answer

Centre: $(-5,4)$

Radius: $9$

Swipe left or right to see the whole diagram →

Completing the square reveals centre (-5, 4) and radius 9.

The sign of radius squared matters

$r^2>0$ : a real circle.
$r^2=0$ : one point at the centre.
$r^2<0$ : no real points satisfy the equation.

Common mistake

(x+5)^2+(y-4)^2=81

, the centre is

(-5,4)

, not

(5,-4)

, and the radius is 9, not 81.

Keep a parameter symbolic

Your turn9709/12 Oct/Nov 2024 Q8(a)

Express $x^2+y^2+px+2y+q=0$ in centre-radius form. State the centre and $r^2$ .

Show worked answer

Answer

$\left(x+\frac p2\right)^2-\frac{p^2}{4}+(y+1)^2-1+q=0$

$\left(x+\frac p2\right)^2+(y+1)^2=\frac{p^2}{4}-q+1$

Centre $\left(-\frac p2,-1\right)$ , $r^2=\frac{p^2}{4}-q+1$ .

PreviousThe Equation of a Circle NextCircles Through Three Points