Algebra

Equation of a circle

Q: What is the equation of a circle in GCSE Maths?

For a circle centred at the origin with radius r, it is x^2 + y^2 = r^2. For Edexcel GCSE you only need circles centred at the origin.

Q: How do you find the equation of a tangent to a circle?

Find the gradient of the radius to the point, flip it and change its sign to get the tangent gradient, then use y - b = m(x - a).

10 past-paper questions, easiest first. Try each one, then reveal the answer.

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Equation of a circle: what you need to know

A circle centred at the origin has a neat equation that comes straight from Pythagoras. Exam questions ask you to find the radius, check points, or find the equation of a tangent.

Key facts

Circle, centre $(0, 0)$ , radius $r$ : $x^2 + y^2 = r^2$ .
So the radius is the square root of the number on the right: $x^2 + y^2 = 36$ has $r = 6$ .
A point is on the circle if its coordinates fit the equation. Substitute to find an unknown: $(3p, p)$ on $x^2 + y^2 = 40$ gives $9p^2 + p^2 = 40$ , so $p = \pm 2$ . Through $(6, 8)$ , the circle is $x^2 + y^2 = 100$ .
A tangent is perpendicular to the radius at the point where it touches.
Perpendicular gradients multiply to $-1$ : if the radius has gradient $m$ , the tangent has gradient $-\frac{1}{m}$ .

Equation of a tangent at a point $(a, b)$

Find the gradient of the radius from $(0, 0)$ to $(a, b)$ : $\frac{b}{a}$ .
Take the negative reciprocal to get the tangent's gradient.
Use $y - b = m(x - a)$ and rearrange to the form asked for.
If the point is on an axis, the tangent is horizontal or vertical: at $(0, 5)$ it is $y = 5$ .

Worked examples

Example 1 Higher · grade 6

A circle has equation $x^2 + y^2 = 49$ . (a) Write down its radius. (b) Is the point $(4, 5)$ inside, on or outside the circle?

Show the solution

(a) $r = \sqrt{49} = 7$
(b) $4^2 + 5^2 = 16 + 25 = 41$
$41 < 49$ , so the point is inside the circle.

Example 2 Higher · grade 8

Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, -4)$ . Give it in the form $ax + by = c$ .

Show the solution

Check: $3^2 + (-4)^2 = 25$ , so the point is on the circle.
Gradient of radius $= \frac{-4}{3}$
Gradient of tangent $= \frac{3}{4}$
$y + 4 = \frac{3}{4}(x - 3)$
$4y + 16 = 3x - 9$
$3x - 4y = 25$

Common mistakes

Halving the number instead of square rooting it: $x^2 + y^2 = 36$ has radius 6, not 18.
Using the radius's gradient as the tangent's gradient. Always take the negative reciprocal.
Writing $3p^2$ when you mean $(3p)^2 = 9p^2$ when substituting a point.

Exam tips

Show the radius gradient and the tangent gradient as separate lines of working: mark schemes usually give a method mark for each, even if the final equation slips.
Any equivalent form is usually accepted, but if the question asks for $ax + by = c$ , clear the fractions so $a$ , $b$ and $c$ are integers.

Questions people ask

What is the equation of a circle in GCSE Maths?

For a circle centred at the origin with radius $r$ , it is $x^2 + y^2 = r^2$ . For Edexcel GCSE you only need circles centred at the origin.

How do you find the equation of a tangent to a circle?

Find the gradient of the radius to the point, flip it and change its sign to get the tangent gradient, then use $y - b = m(x - a)$ .

Exam questions: Equation of a circle (10, easiest first)

Equation of a circle1 markGrade 6CalculatorHigher

November 2018 · Paper 2 Higher · Q15

Find the radius of a circle from its equation.

Show answer

Answer	$6.5$	1 mark

Worked solution Checked against the mark scheme

$x^2 + y^2 = r^2$ , so $r^2 = 42.25$

$r = \sqrt{42.25} = 6.5$

Watch out: Halving 42.25 instead of square rooting.

Equation of a circle4 marksGrade 8No calculatorHigher

November 2024 · Paper 1 Higher · Q20

Find the equation of the tangent to a circle centred at the origin at a point with an unknown coordinate.

Show answer

Answer	$y + \sqrt{3}x = 4$	4 marks

Worked solution Checked against the mark scheme

$p^2 + 1^2 = 4$ , so $p = \sqrt{3}$ (as $p > 0$ )

Radius gradient $= \frac{1}{\sqrt{3}}$ , so tangent gradient $= -\sqrt{3}$

$y - 1 = -\sqrt{3}(x - \sqrt{3})$ , so $y = -\sqrt{3}x + 3 + 1$

$y + \sqrt{3}x = 4$

Watch out: Using the radius gradient as the tangent gradient.

Equation of a circle5 marksGrade 8No calculatorHigher

November 2023 · Paper 1 Higher · Q21

Draw the circle centred at the origin with radius 13 and use a straight line to estimate where they cross.

Show answer

(a)	Circle with centre $(0, 0)$ and radius 13	2 marks
(b)	$x = 7.2, y = 10.8$ and $x = -7.2, y = -10.8$	3 marks

Worked solution Checked against the mark scheme

(a) $x^2 + y^2 = 169$ is a circle centred at the origin with radius $\sqrt{169} = 13$ .

(b) Draw $y = 1.5x$ through $(0, 0)$ , $(4, 6)$ and $(8, 12)$ .

It crosses the circle at about $(7.2, 10.8)$ and $(-7.2, -10.8)$ .

Check: $x^2(1 + 2.25) = 169$ gives $x^2 = 52$ , $x \approx 7.21$

Watch out: Using 169 as the radius, or giving x values without matching y values.

Equation of a circle4 marksGrade 8CalculatorHigher

June 2023 · Paper 3 Higher · Q23

Decide whether the tangent to a circle at a given point passes through a stated point on the x-axis.

Show answer

Answer	No: the tangent is $y = \frac{3}{4}x + \frac{25}{4}$ , which meets the $x$ -axis at $(-\frac{25}{3}, 0)$ , not $(-8, 0)$	4 marks

Worked solution Checked against the mark scheme

Gradient of radius $OP = \dfrac{4}{-3} = -\dfrac{4}{3}$

Tangent is perpendicular, so its gradient is $\dfrac{3}{4}$ .

Tangent: $y - 4 = \frac{3}{4}(x + 3)$ , so $y = \frac{3}{4}x + \frac{25}{4}$

At $x = -8$ : $y = -6 + \frac{25}{4} = \frac{1}{4} \neq 0$

So the tangent does not pass through $(-8, 0)$ : Alex is wrong.

Watch out: Using the gradient of the radius instead of its negative reciprocal.

Equation of a circle4 marksGrade 8No calculatorHigher

June 2022 · Paper 1 Higher · Q20

Find the equation of the tangent to a circle at a given point, using the centre and perpendicular gradients.

Show answer

Answer	$7x + 5y - 82 = 0$	4 marks

Worked solution Checked against the mark scheme

Gradient of radius from $(-1, 3)$ to $(6, 8)$ : $\frac{8-3}{6+1} = \frac{5}{7}$

Tangent is perpendicular: gradient $-\frac{7}{5}$

$y - 8 = -\frac{7}{5}(x - 6)$

$5y - 40 = -7x + 42$ , so $7x + 5y - 82 = 0$

Watch out: Using the gradient of the radius (5/7) as the tangent's gradient.

Equation of a circle4 marksGrade 8CalculatorHigher

June 2019 · Paper 3 Higher · Q22

Use a tangent and a 30° angle to find a circle's radius, then find p for a point (3p, p) on the circle.

Show answer

Answer	$p = 2.5$ (accept 2.5 to 2.53)	4 marks

Worked solution Checked against the mark scheme

Tangent meets radius at 90°, so triangle $OAB$ is right-angled at $A$ .

$OA = 16 \sin 30° = 8$ , so the circle is $x^2 + y^2 = 64$

$(3p)^2 + p^2 = 64 \Rightarrow 10p^2 = 64 \Rightarrow p^2 = 6.4$

$p = 2.529... = 2.5$ (1 d.p.)

Watch out: Writing 3p² instead of (3p)², giving 4p² = 64.

Equation of a circle3 marksGrade 8CalculatorHigher

Specimen · Paper 2 Higher · Q23

Find the equation of the tangent to a circle centred at the origin at a given point.

Show answer

Answer	$y = -\frac{4}{3}x + \frac{25}{3}$	3 marks

Worked solution Checked against the mark scheme

Gradient of radius $OP = \frac{3}{4}$

The tangent is perpendicular, so its gradient is $-\frac{4}{3}$

$3 = -\frac{4}{3}(4) + c \Rightarrow c = \frac{25}{3}$

$y = -\frac{4}{3}x + \frac{25}{3}$

Watch out: Using the radius gradient $\frac{3}{4}$ as the tangent gradient.

Equation of a circle5 marksGrade 9No calculatorHigher

June 2024 · Paper 1 Higher · Q23

Use a circle through a given point and a line through two points on it to find the line's y-intercept.

Show answer

Answer	$k = -5$	5 marks

Worked solution Checked against the mark scheme

Radius $^2 = 5^2 + 10^2 = 125$ , so C is $x^2 + y^2 = 125$

At Q, $(-2)^2 + y^2 = 125 \Rightarrow y^2 = 121 \Rightarrow y = \pm 11$

The gradient of L is positive, so Q is $(-2, -11)$ (using $(-2, 11)$ gives a negative gradient).

Gradient $= \frac{10 - (-11)}{5 - (-2)} = \frac{21}{7} = 3$

$10 = 3(5) + k \Rightarrow k = -5$

Watch out: Taking $y = 11$ for Q without checking the gradient is positive.

Equation of a circle4 marksGrade 9CalculatorHigher

November 2022 · Paper 2 Higher · Q24

Find the equation of the tangent to a circle at a given point, in the form ax + by = c.

Show answer

Answer	$6x + 8y = 35$	4 marks

Worked solution Checked against the mark scheme

Gradient of radius $OP = \frac{2.8}{2.1} = \frac{4}{3}$

Tangent gradient $= -\frac{3}{4}$

$y - 2.8 = -\frac{3}{4}(x - 2.1)$

Multiply by 4: $4y - 11.2 = -3x + 6.3$ , so $3x + 4y = 17.5$

Double for integers: $6x + 8y = 35$

Watch out: Using the radius gradient 4/3 as the tangent gradient.

Equation of a circle5 marksGrade 9CalculatorHigher

November 2020 · Paper 2 Higher · Q22

Find the equation of a circle centred at the origin, given two points on one of its tangents.

Show answer

Answer	$x^2 + y^2 = 80$	5 marks

Worked solution Checked against the mark scheme

Tangent gradient $= \frac{10 - 0}{0 - (-20)} = \frac{1}{2}$ , so the tangent is $y = \frac{1}{2}x + 10$ .

The radius to the point of contact is perpendicular, gradient $-2$ , through the origin: $y = -2x$ .

$\frac{1}{2}x + 10 = -2x$ gives $x = -4$ , $y = 8$ .

$r^2 = (-4)^2 + 8^2 = 80$ , so the circle is $x^2 + y^2 = 80$ .

Watch out: Taking the radius as the distance from the origin to one of the given points (e.g. 20).

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

November 2021 · Paper 1 Higher · Q20: Draw a straight line on a circle graph to estimate where the line and circle meet.

Equation of a circle

Equation of a circle: what you need to know

Key facts

Equation of a tangent at a point (a,b)(a, b)(a,b)

Worked examples

Common mistakes

Exam tips

Questions people ask

Exam questions: Equation of a circle (10, easiest first)

Also tested in these questions

Equation of a tangent at a point $(a, b)$