Is x^2 + y^2 = 10 a circle?

Yes, with radius 10. The radius does not have to be a whole number.

J560 · 7.01–7.04GraphsHigherGrades 7–9

Equation of a Circle

Q: What about circles not centred at the origin?

They have equation (x - a)^2 + (y - b)^2 = r^2, but GCSE J560 only requires circles centred at the origin; the general form is A-Level.

Q: Why is the tangent perpendicular to the radius?

It is a circle theorem: the radius to the point of contact meets the tangent at a right angle.

A circle is the set of all points a fixed distance from its centre. For a circle centred at the origin, that single fact, combined with Pythagoras’ theorem, gives its equation: $x^{2} + y^{2} = r^{2}$ . The other skill OCR tests is finding the equation of a $\textbf{tangent}$ — the line that touches the circle at one point.

40 min Graphs

The big picture

Take any point $(x, y)$ on a circle of radius $r$ centred at the origin. Drop a vertical line to the $x$ -axis and you have a right-angled triangle with legs $x$ and $y$ and hypotenuse $r$ . Pythagoras gives $x^{2} + y^{2} = r^{2}$ immediately.

The tangent at a point is perpendicular to the radius drawn to that point — a circle theorem. So finding a tangent combines three skills: the gradient of the radius, the perpendicular gradient, and the equation of a line through a point.

Equation of a Circle - CorbettmathsCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Before you start

Try a short calculation from the preceding topic without the notes.
Check the tier and grade band shown in this lesson.

Watch, pause, try

Video by corbettmaths. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Recognise and use the equation $x^{2} + y^{2} = r^{2}$

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: What is the radius of $x^{2} + y^{2} = 81$ ?

Show answer and reasoning

$9$ . $\sqrt{81}$ .

After watching 2: Which is the equation of a circle, centre the origin, radius 4?

Show answer and reasoning

$x^{2} + y^{2} = 16$ . $r^{2} = 16$ .

What you'll learn

Recognise and use the equation $x^{2} + y^{2} = r^{2}$
Find the radius from an equation, and check whether a point lies on a circle
Find the equation of the tangent to a circle at a given point
Solve problems involving circles and straight lines

The equation x² + y² = r²

A circle centred at the origin with radius $r$ has equation $x^{2} + y^{2} = r^{2}$ .

The right-hand side is $r^{2}$ , not $r$ : the circle $x^{2} + y^{2} = 49$ has radius 7.

A point lies on the circle if its coordinates satisfy the equation. If $x^{2} + y^{2} < r^{2}$ the point is inside; if greater, outside.

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

Circle, centre (0, 0), radius r.

Example 1Does the point

(-5, 12)

lie on the circle

x^{2} + y^{2} = 169

(-5)^{2} + 12^{2} = 25 + 144 = 169

AnswerYes — it satisfies the equation, so it lies on the circle of radius 13.

Your turnSame method, new numbers — try it before moving on.

Find the exact radius of the circle $x^2 + y^2 = 60$ , in simplified surd form.

Your answer

Type √ as sqrt, e.g. 3sqrt5.

Tangent at a point

Step 1: find the gradient of the radius from $(0, 0)$ to the point $(a, b)$ : $\dfrac{b}{a}$ .

Step 2: the tangent is perpendicular, so its gradient is $-\dfrac{a}{b}$ .

Step 3: use $y - b = m(x - a)$ , or substitute into $y = mx + c$ , to find the equation.

Example 1Find the equation of the tangent to

x^{2} + y^{2} = 25

at the point

(3, 4)

1Gradient of the radius:

\dfrac{4}{3}

2Tangent gradient:

-\dfrac{3}{4}

y - 4 = -\tfrac34(x - 3)

, so

4y - 16 = -3x + 9

Answer

3x + 4y = 25

(or

y = -\tfrac34 x + \tfrac{25}{4}

)

Your turnSame method, new numbers — try it before moving on.

Find the equation of the tangent to $x^2 + y^2 = 13$ at $(2, 3)$ .

Your answer

Type it like y = -2/3x + 13/3.

Why this works

For the circle $x^{2} + y^{2} = r^{2}$ , the tangent at $(a, b)$ always simplifies to $ax + by = r^{2}$ . Here, $3x + 4y = 25$ — a neat check on your working.

Where lines meet circles

To find where a line crosses a circle, substitute the line’s equation into the circle’s and solve the resulting quadratic.

Two solutions: the line cuts the circle twice. One repeated solution: the line is a tangent. No real solutions: the line misses the circle.

Example 1Find where the line

y = x - 1

meets the circle

x^{2} + y^{2} = 13

1Substitute:

x^{2} + (x - 1)^{2} = 13

, so

2x^{2} - 2x + 1 = 13

2x^{2} - 2x - 12 = 0

, so

x^{2} - x - 6 = 0

and

(x - 3)(x + 2) = 0

x = 3

gives

y = 2

;

x = -2

gives

y = -3

Answer

(3, 2)

and

(-2, -3)

Your turnSame method, new numbers — try it before moving on.

The line $y = 2x$ meets the circle $x^2 + y^2 = 20$ at two points. Find their $x$ -coordinates.

Your answer

Separate with a comma.

Tip — Check each intersection point in the circle’s equation: $9 + 4 = 13$ ✓ and $4 + 9 = 13$ ✓.

Think like an examiner

State the gradient of the radius and the perpendicular gradient separately.
Give the tangent equation in the form requested.
For intersections, show the substituted quadratic and pair each $x$ with its $y$ .

Circles

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

Circle at the origin.

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

Tangent ⊥ radius.

ax + by = r^{2}

Tangent at (a, b).

Watch out for these

Reading the radius of

x^{2} + y^{2} = 36

as 36.

The radius is

\sqrt{36} = 6

Using the radius gradient as the tangent gradient.

The tangent is perpendicular: use the negative reciprocal.

Expanding

(x - 1)^{2}

x^{2} + 1

(x - 1)^{2} = x^{2} - 2x + 1

Pairing

x = 3

with

y = -3

Substitute each

x

into the line to get its own

y

Stretch yourself

The line $y = x + c$ is a tangent to the circle $x^{2} + y^{2} = 8$ . Find the possible values of $c$ and the point of contact for the positive value.

Hint — Substitute to get a quadratic in $x$ . For a tangent, the quadratic has one repeated solution, so its discriminant is zero.

Questions students ask

What about circles not centred at the origin?

They have equation

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

, but GCSE J560 only requires circles centred at the origin; the general form is A-Level.

Why is the tangent perpendicular to the radius?

It is a circle theorem: the radius to the point of contact meets the tangent at a right angle.

x^{2} + y^{2} = 10

a circle?

Yes, with radius

\sqrt{10}

. The radius does not have to be a whole number.

Key takeaways

$x^{2} + y^{2} = r^{2}$ is a circle centre $(0, 0)$ , radius $r$ .
Substitute a point to test whether it lies on the circle.
Tangent gradient is the negative reciprocal of the radius gradient.
The tangent at $(a, b)$ is $ax + by = r^{2}$ .
Substitute a line into the circle to find intersections.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 62 marks

The point $(6, k)$ lies on the circle $x^2 + y^2 = 100$ , and $k > 0$ . Find $k$ .[2]

Question 2Grade 83 marks

Find the equation of the tangent to $x^2 + y^2 = 20$ at the point $(-4, 2)$ .[3]

Question 3Grade 94 marks

Show that the line $y = x + 4$ meets the circle $x^2 + y^2 = 8$ at exactly one point, and find its coordinates.[4]

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

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OCR J560 official specification (opens in a new tab)
Check the assessed scope and practical or mathematical requirements.
Read or explore this topic (opens in a new tab)
An external explanation or interactive activity to use alongside the lesson. It may include extension material.

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

Which point lies on $x^{2} + y^{2} = 50$ ?
What is the gradient of the tangent to $x^{2} + y^{2} = 20$ at $(4, 2)$ ?
Find the tangent to $x^{2} + y^{2} = 10$ at $(1, 3)$ .
How many times does $y = 6$ meet $x^{2} + y^{2} = 25$ ?

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 5 lessons

Equation of a Circle

A circle is the set of all points a fixed distance from its centre. For a circle centred at the origin, that single fact, combined with Pythagoras’ theorem, gives its equation:

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

. The other skill OCR tests is finding the equation of a

\textbf{tangent}

— the line that touches the circle at one point.

40 min Graphs

Watch, pause, try

Video by corbettmaths. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Recognise and use the equation

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

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: What is the radius of $x^{2} + y^{2} = 81$ ?

Show answer and reasoning

$9$ . $\sqrt{81}$ .

After watching 2: Which is the equation of a circle, centre the origin, radius 4?

Show answer and reasoning

$x^{2} + y^{2} = 16$ . $r^{2} = 16$ .

Practise it the way it is examined

Question 1Grade 62 marks

The point $(6, k)$ lies on the circle $x^2 + y^2 = 100$ , and $k > 0$ . Find $k$ .[2]

Question 2Grade 83 marks

Find the equation of the tangent to $x^2 + y^2 = 20$ at the point $(-4, 2)$ .[3]

Question 3Grade 94 marks

Show that the line $y = x + 4$ meets the circle $x^2 + y^2 = 8$ at exactly one point, and find its coordinates.[4]

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

Which point lies on

x^{2} + y^{2} = 50

What is the gradient of the tangent to

x^{2} + y^{2} = 20

(4, 2)

Find the tangent to

x^{2} + y^{2} = 10

(1, 3)

How many times does

y = 6

meet

x^{2} + y^{2} = 25