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

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

1MA1 · A16AlgebraHigherGrades 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 1MA1 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 Edexcel 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.

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.

Show worked solution

Solution

1 $r = \sqrt{60} = \sqrt{4 \times 15}$ .
2 $= 2\sqrt{15}$ .

Answer: $2\sqrt{15}$

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.

Show worked solution

Solution

1Radius gradient $\frac{3}{2}$ , so the tangent gradient is $-\frac{2}{3}$ .
2 $3 = -\frac{2}{3}(2) + c$ , so $c = \frac{13}{3}$ .

Answer: $y = -\frac{2}{3}x + \frac{13}{3}$

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.

Show worked solution

Solution

1 $x^2 + 4x^2 = 20$ , so $5x^2 = 20$ .
2 $x^2 = 4$ , so $x = \pm 2$ .

Answer: $x = 2$ or $x = -2$

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.

Reveal solution

SolutionSubstituting:

x^{2} + (x + c)^{2} = 8

, so

2x^{2} + 2cx + c^{2} - 8 = 0

. For a tangent, the discriminant is zero:

(2c)^{2} - 4(2)(c^{2} - 8) = 0

, so

4c^{2} - 8c^{2} + 64 = 0

, giving

c^{2} = 16

and

c = \pm 4

. For

c = 4

2x^{2} + 8x + 8 = 0

, i.e.

(x + 2)^{2} = 0

, so

x = -2

and

y = 2

. The point of contact is

(-2, 2)

. Check:

4 + 4 = 8

✓.

\blacksquare

Questions students ask

What about circles not centred at the origin?

They have equation

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

, but GCSE 1MA1 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]
Show mark scheme
- M1 $36 + k^2 = 100$
- A1 $k = 8$
Answer: $k = 8$

Question 2Grade 83 marks

Find the equation of the tangent to $x^2 + y^2 = 20$ at the point $(-4, 2)$ .[3]
Show mark scheme
- M1Radius gradient $-\frac{1}{2}$
- M1Tangent gradient $2$ used with $(-4, 2)$
- A1 $y = 2x + 10$
Answer: $y = 2x + 10$

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]
Show mark scheme
- M1 $x^2 + (x + 4)^2 = 8$
- M1 $x^2 + 4x + 4 = 0$
- A1 $(x + 2)^2 = 0$ : one repeated root
- A1 $(-2, 2)$
Answer: $(-2, 2)$

Unlimited practice

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