GCSE Maths · Algebra

Equation of a Circle GCSE Questions and Worked Answers

A circle centred at the origin with radius r has equation x² + y² = r². The equation comes from Pythagoras: every point (x, y) on the circle is exactly r units from (0, 0).

Edexcel · AQA · OCRHigher tier15 original questions

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Begin with distance from the centre

What you need to know about the equation of a circle

A circle is the set of all points the same distance from one centre. Put the centre at the origin O(0, 0), choose a point P(x, y) on the circle and call the fixed distance OP the radius r. The horizontal and vertical movements to P are x and y, so they form a right-angled triangle with OP.

Coordinates → distance → circle equation

Build x² + y² = r² from one point

For P(3, 4), the horizontal and vertical distances are 3 and 4. Pythagoras gives the distance from the origin.

CoordinatesP(3, 4)move 3 across and 4 up

Pythagoras3² + 4² = r²the radius is the hypotenuse

Fixed distancer² = 25every point on this circle satisfies x² + y² = 25

The radius, horizontal distance and vertical distance form a right-angled triangle.

Substitution calculates the point’s squared distance d² = x² + y², which you compare directly with r².

A 90° turn swaps the horizontal and vertical changes and reverses one direction. That is why the perpendicular gradient is the negative reciprocal.

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

State the centre and radius

What the problem asks: Read the centre and radius from an equation such as x² + y² = 81.

How to solve it: The centre is (0, 0). Since the right side is r², take its positive square root to find r.

$x^2+y^2=81\quad\Rightarrow\quad r=9$

Write a circle equation

What the problem asks: Write the equation of a circle centred at the origin when its radius or a point on it is given.

How to solve it: Find r², then substitute it into x² + y² = r².

$r=6\quad\Rightarrow\quad x^2+y^2=36$

Test whether a point lies on a circle

What the problem asks: Decide whether a coordinate is on, inside or outside a stated circle.

How to solve it: Substitute its x- and y-values. Compare x² + y² with r²: equal means on, smaller means inside and larger means outside.

$x^2+y^2\begin{cases}<r^2&\text{inside}\\=r^2&\text{on}\\>r^2&\text{outside}\end{cases}$

Find a missing coordinate

What the problem asks: One coordinate of a point on the circle is missing.

How to solve it: Substitute the known coordinate, solve for the square of the missing value and use the point’s position to choose the positive or negative root.

$y^2=r^2-x^2\quad\Rightarrow\quad y=\pm\sqrt{r^2-x^2}$

Find the tangent at a point

What the problem asks: Find the equation of the line touching the circle at one stated point.

How to solve it: Find the gradient of the radius to that point. The tangent is perpendicular, so use the negative reciprocal gradient and the coordinates of the touching point.

$m_{\text{radius}}m_{\text{tangent}}=-1$

A reliable routine

Method for a tangent to a circle centred at the origin

Use this method when the circle, the touching point and a tangent equation are involved. It works because a tangent is perpendicular to the radius at the point of contact.

Check that the stated point lies on the circle by substituting its coordinates.
Find the gradient of the radius from (0, 0) to the touching point.
Use the negative reciprocal for the perpendicular tangent gradient.
Substitute the tangent gradient and touching point into y − y₁ = m(x − x₁).
Rearrange into the requested form and check that the line passes through the touching point.

Check: A vertical radius has an undefined gradient and a horizontal tangent; a horizontal radius has gradient 0 and a vertical tangent.

Fully worked

Equation of a circle GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Read the centre and radius

2 marks

Question

State the centre and radius of

$x^2+y^2=64$

The equation has centre $(0,0)$ and right side $r^2$ .

$r^2=64$

$r=8$

$\boxed{\text{centre }(0,0),\ \text{radius }8}$

Example 2

Write the equation from a radius

2 marks

Question

Write the equation of the circle centred at the origin with radius $7$ .

Start with $x^2+y^2=r^2$ .

$r^2=7^2=49$

$\boxed{x^2+y^2=49}$

Example 3

Check a point

2 marks

Question

Does the point $(6,8)$ lie on $x^2+y^2=100$ ?

Substitute $x=6$ and $y=8$ .

$6^2+8^2=36+64$

$6^2+8^2=100$

The point satisfies the equation, so

$\boxed{(6,8)\text{ lies on the circle}.}$

Example 4

Find a missing coordinate

3 marks

Question

The point $(4,y)$ lies on $x^2+y^2=25$ and is above the $x$ -axis. Find $y$ .

For Q(4, 3), substitute x = 4; the point is above the x-axis, so choose y = +3 rather than −3.

Substitute $x=4$ .

$4^2+y^2=25$

$16+y^2=25$

$y^2=9$

The point is above the $x$ -axis, so $y$ is positive.

$\boxed{y=3}$

Example 5

Find a tangent equation

4 marks

Question

Find the equation of the tangent to $x^2+y^2=25$ at $(3,4)$ .

The radius from $(0,0)$ to $(3,4)$ has gradient

$m_r=\frac43$

The tangent is perpendicular, so its gradient is the negative reciprocal.

$m_t=-\frac34$

Use the point $(3,4)$ .

$y-4=-\frac34(x-3)$

$4y-16=-3x+9$

$\boxed{3x+4y=25}$

Example 6

Tangent with a negative radius gradient

4 marks

Question

Find the tangent to $x^2+y^2=25$ at $(-4,3)$ .

The radius gradient is

$m_r=\frac{3}{-4}=-\frac34$

The perpendicular tangent gradient is

$m_t=\frac43$

Use $(-4,3)$ .

$y-3=\frac43(x+4)$

$3y-9=4x+16$

$\boxed{3y=4x+25}$

Example 7

Find where a circle meets an axis

3 marks

Question

Find the coordinates where $x^2+y^2=49$ meets the $x$ -axis.

Every point on the $x$ -axis has $y=0$ .

$x^2+0^2=49$

$x=\pm7$

$\boxed{(-7,0)\text{ and }(7,0)}$

Example 8

Handle a horizontal tangent

3 marks

Question

Find the tangent to $x^2+y^2=25$ at $(0,5)$ .

The radius from $(0,0)$ to $(0,5)$ is vertical.

A line perpendicular to a vertical line is horizontal. The horizontal line through $(0,5)$ is

$\boxed{y=5}$

15 original questions · total 49 marks

Equation of a circle GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 50 minutes · show substitutions and keep exact values unless rounding is requested · answers start collapsed

State a radius

1 mark

State the radius of $x^2+y^2=121$ .

Show worked answer

$r=\sqrt{121}=\boxed{11}$

Write an equation

2 marks

Write the equation of the circle centred at $(0,0)$ with radius $4$ .

Show worked answer

$r^2=4^2=16$

$\boxed{x^2+y^2=16}$

Use a point to find the circle

3 marks

A circle centred at the origin passes through $(5,12)$ . Find its equation.

Show worked answer

$r^2=5^2+12^2$

$r^2=25+144=169$

$\boxed{x^2+y^2=169}$

Test a coordinate

2 marks

Show that $(-8,6)$ lies on $x^2+y^2=100$ .

Show worked answer

$(-8)^2+6^2=64+36$

$(-8)^2+6^2=100$

Therefore $\boxed{(-8,6)\text{ lies on the circle}}$ .

Classify a point

3 marks

Is $(2,3)$ inside, on or outside $x^2+y^2=16$ ?

Show worked answer

$2^2+3^2=4+9=13$

Since $13<16$ , the point is nearer to the centre than the radius.

$\boxed{(2,3)\text{ is inside the circle}.}$

Find two possible coordinates

3 marks

The point $(x,8)$ lies on $x^2+y^2=100$ . Find both possible values of $x$ .

Show worked answer

$x^2+8^2=100$

$x^2=36$

$\boxed{x=6\text{ or }x=-6}$

Use a quadrant

3 marks

The point $(x,-5)$ lies on $x^2+y^2=169$ in the third quadrant. Find $x$ .

Show worked answer

$x^2+(-5)^2=169$

$x^2=144$

In the third quadrant, $x$ is negative.

$\boxed{x=-12}$

Find axis intercepts

3 marks

Find all four points where $x^2+y^2=36$ meets the coordinate axes.

Show worked answer

On the $x$ -axis, $y=0$ , so $x=\pm6$ .

On the $y$ -axis, $x=0$ , so $y=\pm6$ .

$\boxed{(6,0),\ (-6,0),\ (0,6),\ (0,-6)}$

Find the point before finding its tangent

5 marks

A point $P$ lies on $x^2+y^2=169$ . Its $x$ -coordinate is $5$ and $y>0$ . Find the equation of the tangent at $P$ .

Show worked answer

First find the missing coordinate.

$5^2+y^2=169$

$y^2=144$

Because $y>0$ , $P=(5,12)$ .

$m_r=\frac{12}{5}$

$m_t=-\frac{5}{12}$

$y-12=-\frac5{12}(x-5)$

$\boxed{5x+12y=169}$

Find where a line meets a circle

5 marks

Find the two points where the line $y=x+1$ meets the circle $x^2+y^2=25$ .

Show worked answer

Substitute $y=x+1$ into the circle equation.

$x^2+(x+1)^2=25$

$2x^2+2x-24=0$

$x^2+x-12=0$

$(x+4)(x-3)=0$

So $x=-4$ or $x=3$ . Using $y=x+1$ gives $y=-3$ or $y=4$ .

$\boxed{(-4,-3)\text{ and }(3,4)}$

Tangent in another quadrant

4 marks

Find the tangent to $x^2+y^2=50$ at $(5,-5)$ .

Show worked answer

$m_r=\frac{-5}{5}=-1$

The perpendicular gradient is $1$ .

$y+5=x-5$

$\boxed{y=x-10}$

Find a vertical tangent

3 marks

Find the tangent to $x^2+y^2=81$ at $(9,0)$ .

Show worked answer

The radius to $(9,0)$ is horizontal. Its perpendicular tangent is vertical and passes through $x=9$ .

$\boxed{x=9}$

Recover the circle from a tangent point

4 marks

A circle centred at the origin has a tangent at $(7,24)$ . Find the circle equation.

Show worked answer

The point lies on the circle, so

$r^2=7^2+24^2$

$r^2=49+576=625$

$\boxed{x^2+y^2=625}$

Find an exact coordinate

3 marks

A point $(3,y)$ lies on $x^2+y^2=20$ above the $x$ -axis. Find $y$ exactly.

Show worked answer

$3^2+y^2=20$

$y^2=11$

Above the axis means $y>0$ .

$\boxed{y=\sqrt{11}}$

Link a tangent and an intercept

5 marks

The tangent to $x^2+y^2=100$ at $(6,8)$ meets the $y$ -axis at $B$ . Find the coordinates of $B$ .

Show worked answer

The radius gradient is

$m_r=\frac86=\frac43$

so the tangent gradient is $-3/4$ .

$y-8=-\frac34(x-6)$

At the $y$ -axis, $x=0$ .

$y-8=-\frac34(0-6)=\frac92$

$y=\frac{25}{2}$

$\boxed{B\left(0,\frac{25}{2}\right)}$

Examiner-style feedback

Common circle equations mistakes

Calling r² the radius

In x² + y² = 49, the radius is √49 = 7, not 49.

Losing a negative coordinate when squaring

Use brackets: (−4)² = 16. The square is positive.

Choosing only one square root

A missing coordinate may have positive and negative values unless a quadrant or diagram selects one.

Using the radius gradient for the tangent

The tangent is perpendicular. For two non-vertical lines, their gradients multiply to −1.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

A circle is a fixed distance from its centre.
At the origin, Pythagoras gives x² + y² = r².
Substitution tests a coordinate.
A tangent is perpendicular to the radius at the touching point.

Quick answers

Equation of a circle FAQ

What is the GCSE equation of a circle?

For a circle centred at the origin, the equation is x² + y² = r², where r is the radius.

Why is the radius squared?

The equation comes from Pythagoras: the squared horizontal and vertical distances add to the squared direct distance.

How do I know if a point lies on the circle?

Substitute its coordinates. It lies on the circle when x² + y² equals r².

How do I find a tangent equation?

Find the radius gradient, take the negative reciprocal for the tangent, then use the touching point in a straight-line equation.

Build connected skills

What to revise next

Right triangles

Trigonometry

Use angles and side ratios in right-angled triangles.

Revise trigonometry

Perpendicular lines

Circle theorems

Use tangent, chord and angle relationships in circles.

Revise circle theorems

Line–circle intersections

Simultaneous equations

Solve linked linear and quadratic equations, including a line meeting a circle.

Revise simultaneous equations

Content standards

Curriculum and rights review

Reviewed 5 September 2026 against DfE content A16, Pearson Edexcel A16, AQA A16 and OCR J560 sections 7.01f and 7.02b. All questions and diagrams are original.

Official specification references