A16AlgebraHigherGrades 7–9

Equation of a Circle

Q: Do I need circles not centred at the origin?

Not for AQA GCSE. The general form (x - a)^2 + (y - b)^2 = r^2 is A-Level.

Q: What if the point has a zero coordinate?

The tangent is horizontal or vertical, for example y = 5 at (0, 5).

Every point on a circle of radius 6 centred at the origin is exactly 6 units from $(0, 0)$ . Pythagoras turns that fact into an equation: $x^{2} + y^{2} = 36$ . From there you can test points, find tangents and work out where lines cross the circle.

45 min AQA GCSE Maths

The big picture

At GCSE, circles are always centred at the origin, so the equation is $x^{2} + y^{2} = r^{2}$ . The main exam skill is the tangent: it meets the radius at 90°, so its gradient is the negative reciprocal of the radius gradient.

Finding where a line meets a circle combines this topic with simultaneous equations. Substituting the line into the circle gives a quadratic, and its solutions are the $x$ -coordinates of the intersection points.

What you'll learn

Recognise and write the equation $x^{2} + y^{2} = r^{2}$
Decide whether a point is inside, on or outside a circle
Find the equation of a tangent at a given point
Find the intersections of a line and a circle

The equation

A point $(x, y)$ on the circle forms a right-angled triangle with legs $x$ and $y$ and hypotenuse $r$ , so $x^{2} + y^{2} = r^{2}$ .

$x^{2} + y^{2} = 36$ is a circle centred at the origin with radius 6. $x^{2} + y^{2} = 20$ has radius $\sqrt{20} = 2\sqrt5$ .

To test a point, substitute it: less than $r^{2}$ is inside, equal is on, greater is outside.

Example 1Is

(4, -2)

inside, on or outside the circle

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

4^{2} + (-2)^{2} = 16 + 4 = 20

20 < 36

AnswerInside the circle

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

Is the point $(5, -3)$ inside, on or outside the circle $x^2 + y^2 = 34$ ?

Your answer

Type inside, on or outside.

Show worked solution

Solution

1 $5^2 + (-3)^2 = 25 + 9 = 34$ .
2This equals $r^2$ , so the point is on the circle.

Answer: on

Tip — The number on the right is $r^{2}$ , not $r$ . Square root it to find the radius.

Tangents

A tangent touches the circle at one point and is perpendicular to the radius there.

Find the gradient of the radius from $(0, 0)$ to the point, take the negative reciprocal, then use $y - y_1 = m(x - x_1)$ .

Example 1Find the equation of the tangent to

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

(-3, 4)

1Gradient of radius:

\dfrac{4}{-3} = -\dfrac43

2Tangent gradient:

\dfrac34

y - 4 = \dfrac34(x + 3)

, so

4y - 16 = 3x + 9

Answer

4y = 3x + 25

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

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

Your answer

Type it like y = -1/2x + 5.

Show worked solution

Solution

1Radius gradient: $\frac{4}{2} = 2$ , so the tangent gradient is $-\frac{1}{2}$ .
2 $4 = -\frac{1}{2}(2) + c$ , so $c = 5$ .

Answer: $y = -\frac{1}{2}x + 5$

Why this works

Check the answer by substituting the point: $4(4) = 3(-3) + 25$ gives $16 = 16$ ✓. A tangent must pass through the point it touches.

Lines meeting circles

Substitute the line into the circle equation, expand, and solve the quadratic.

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

Example 1Find where

y = 2x + 5

meets

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

x^{2} + (2x + 5)^{2} = 10

5x^{2} + 20x + 15 = 0

, so

x^{2} + 4x + 3 = 0

(x + 1)(x + 3) = 0

x = -1

x = -3

4Substitute into the line:

y = 3

y = -1

Answer

(-1, 3)

and

(-3, -1)

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

The line $y = x + 1$ meets the circle $x^2 + y^2 = 25$ at two points. Find their $x$ -coordinates.

Your answer

Separate with a comma.

Show worked solution

Solution

1 $x^2 + (x + 1)^2 = 25$ gives $2x^2 + 2x - 24 = 0$ .
2 $x^2 + x - 12 = 0$ , so $(x + 4)(x - 3) = 0$ .

Answer: $x = 3$ or $x = -4$

Tip — Always substitute back into the linear equation to find $y$ — the circle equation gives two $y$ -values for each $x$ .

Think like an examiner

Show the gradient of the radius before the tangent gradient.
Give the tangent equation in the form asked for.
Pair each $x$ -value with its $y$ -value for intersections.

Remember these

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

Circle centre (0, 0), radius r.

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

Tangent is perpendicular to the radius.

y - y_1 = m(x - x_1)

Line through a point with gradient m.

Watch out for these

Stating the radius of

x^{2} + y^{2} = 49

as 49.

The radius is 7.

Using the radius gradient for the tangent.

Take the negative reciprocal.

Writing

(2x + 5)^{2} = 4x^{2} + 25

4x^{2} + 20x + 25

Finding

y

from the circle and getting extra points.

Substitute into the line.

Stretch yourself

The tangent to $x^{2} + y^{2} = 20$ at the point $(2, -4)$ meets the axes at $A$ and $B$ . Find the area of triangle $OAB$ .

Hint — Find the tangent, then its $x$ - and $y$ -intercepts.

Reveal solution

SolutionRadius gradient:

\dfrac{-4}{2} = -2

, so the tangent gradient is

\dfrac12

. Tangent:

y + 4 = \dfrac12(x - 2)

, giving

y = \dfrac12 x - 5

, or

x - 2y = 10

. Intercepts:

A(10, 0)

and

B(0, -5)

. Area

= \dfrac12 \times 10 \times 5 = 25

square units.

\blacksquare

Questions students ask

Do I need circles not centred at the origin?

Not for AQA GCSE. The general form

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

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 90°.

What if the point has a zero coordinate?

The tangent is horizontal or vertical, for example

y = 5

(0, 5)

Key takeaways

$x^{2} + y^{2} = r^{2}$ is a circle centred at the origin.
Substitute a point to test inside, on or outside.
Tangent gradient is the negative reciprocal of the radius gradient.
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

A circle has centre $(0, 0)$ and passes through $(-5, 12)$ . Write down its equation.[2]
Show mark scheme
- M1Radius $\sqrt{25 + 144} = 13$ or $r^2 = 169$
- A1 $x^2 + y^2 = 169$
Answer: $x^2 + y^2 = 169$

Question 2Grade 83 marks

Find the equation of the tangent to the circle $x^2 + y^2 = 45$ at the point $(3, -6)$ .[3]
Show mark scheme
- M1Gradient of radius $= -2$
- M1Tangent gradient $= \frac{1}{2}$
- A1 $y = \frac{1}{2}x - \frac{15}{2}$ or $2y = x - 15$
Answer: $2y = x - 15$

Question 3Grade 94 marks

Show that the line $y = x - 10$ is a tangent to the circle $x^2 + y^2 = 50$ , and find the point where it touches.[4]
Show mark scheme
- M1Substitutes: $x^2 + (x - 10)^2 = 50$
- M1Simplifies to $x^2 - 10x + 25 = 0$
- A1 $(x - 5)^2 = 0$ has one repeated root, so the line is a tangent
- A1Point $(5, -5)$
Answer: $(5, -5)$

Unlimited practice

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