Equation of a Circle

Q: How do I find the centre and radius from an expanded circle equation?

Group the $x$ and $y$ terms, complete the square for each, and move the constants to the right. The right-hand side is $r^2$.

Q: How do I check whether a point is inside a circle?

Substitute the point into the left side of $(x - a)^2 + (y - b)^2$. Less than $r^2$ means inside; more means outside.

Q: Do I need completing the square in Grade 12?

Yes. You need it to rewrite circle equations in centre-radius form, and you may also use it to solve quadratics.

The equation of a circle comes straight from the distance formula. In Paper 2 you will need to write equations, find the centre and radius by completing the square, and test whether points or lines meet the circle.

Circle with centre at the origin

Every point $(x; y)$ on a circle with centre $(0; 0)$ and radius $r$ is a distance $r$ from the origin:

$x^{2} + y^{2} = r^{2}$

Example. A circle with centre at the origin passes through $(- 5; 12)$ :

$r^{2} = (- 5)^{2} + 1 2^{2} = 169 \Rightarrow x^{2} + y^{2} = 169$

To test whether a point lies on, inside or outside the circle, substitute it into $x^{2} + y^{2}$ and compare with $r^{2}$ .

Circle with any centre

A circle with centre $(a; b)$ and radius $r$ :

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

If the equation is given in expanded form, complete the square for $x$ and for $y$ .

Example. $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$

$(x^{2} - 6 x + 9) + (y^{2} + 4 y + 4) = 12 + 9 + 4$

$(x - 3)^{2} + (y + 2)^{2} = 25$

Centre $(3; - 2)$ and radius 5.

For the equation to be a circle, the right-hand side after completing the square must be positive.

Diameter given. The centre is the midpoint of the diameter and the radius is half its length.

Key terms

Radius: The distance from the centre to any point on the circle.
Completing the square: Rewriting $x^{2} + b x$ as $(x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}$ .
Diameter: A chord through the centre; its midpoint is the centre.

Common mistakes

Reading the centre of $(x - 3)^{2} + (y + 2)^{2} = 25$ as $(- 3; 2)$ .
Giving the radius as 25 instead of $25 = 5$ .
Forgetting to add the completed-square constants to both sides.
Not dividing through when $x^{2}$ and $y^{2}$ have a coefficient other than 1.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Determine the equation of the circle with centre at the origin that passes through $(- 5; 12)$ .

Worked solution

Hidden - try the question first

Solution

$x^{2} + y^{2} = r^{2}$
$r^{2} = 25 + 144 = 169$
$x^{2} + y^{2} = 169$

Question 2

5 marks total

routine

Determine the centre and radius of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .

Worked solution

Hidden - try the question first

Solution

$x^{2} + y^{2} = r^{2}$
$(x - 3)^{2} + (y + 2)^{2} = 25$
Centre $(3; - 2)$
Radius 5

Question 3

7 marks total

complex

$A (- 1; 3)$ and $B (5; - 5)$ are the endpoints of a diameter of a circle.

(a) Determine the equation of the circle. (4)

(b) Show that $P (6; 2)$ lies on the circle. (2)

Worked solution

Hidden - try the question first

Solution

(a) Centre $= (\frac{- 1 + 5}{2}; \frac{3 - 5}{2}) = (2; - 1)$
$r^{2} = (2 + 1)^{2} + (- 1 - 3)^{2} = 25$
$(x - 2)^{2} + (y + 1)^{2} = 25$
(b) $(6 - 2)^{2} + (2 + 1)^{2} = 16 + 9$
$= 25$ , so $P$ lies on the circle
(c) $90°$ ( $∠$ in semi-circle)

Question 4

9 marks total

problem-solving

Given $x^{2} + y^{2} + 4 x - 2 y + k = 0$ .

(a) For which values of $k$ does this equation represent a circle? (3)

(b) Let $k = - 3$ . Determine the coordinates of the points where the line $y = x + 3$ cuts the circle. (4)

Worked solution

Hidden - try the question first

Solution

(a) $= (\frac{- 1 + 5}{2}; \frac{3 - 5}{2}) = (2; - 1)$
A circle needs $r^{2} > 0$ : $5 - k > 0$
$k < 5$
(b) $(x + 2)^{2} + (y - 1)^{2} = 8$
Substitute $y = x + 3$ : $(x + 2)^{2} + (x + 2)^{2} = 8$
$(x + 2)^{2} = 4 \Rightarrow x = 0$ or $x = - 4$
$r$ and $(- 4; - 1)$
(c) $(0 + 4)^{2} + (3 + 1)^{2} = 32$
$= 42$ units

FAQs

How do I find the centre and radius from an expanded circle equation?

Group the $x$ and $y$ terms, complete the square for each, and move the constants to the right. The right-hand side is $r^{2}$ .

How do I check whether a point is inside a circle?

Substitute the point into the left side of $(x - a)^{2} + (y - b)^{2}$ . Less than $r^{2}$ means inside; more means outside.

Do I need completing the square in Grade 12?

Yes. You need it to rewrite circle equations in centre-radius form, and you may also use it to solve quadratics.

Circle with centre at the origin

Circle with any centre

Key terms

Common mistakes

Practice questions

FAQs

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