Algebra

Functions and the Equation of a Circle (Higher Tier)

Year 10 · Year 11

✓By the end of this lesson students will be able to understand and use function notation f(x).
✓By the end of this lesson students will be able to find and evaluate composite functions fg(x).
✓By the end of this lesson students will be able to find and evaluate inverse functions f⁻¹(x).
✓By the end of this lesson students will be able to understand and use the equation of a circle x² + y² = r² centred at the origin.
✓By the end of this lesson students will be able to solve problems involving functions and the equation of a circle.

Key concepts

Function Notation

A function is a rule that assigns each input value (usually 'x') to exactly one output value (usually 'y'). Function notation, f(x), is used to represent the output of a function when the input is x. For example, if f(x) = 2x + 1, then f(3) means substitute x=3 into the function, so f(3) = 2(3) + 1 = 7. The 'f' is the name of the function, and 'x' is the input variable.

Composite Functions

A composite function is a function formed by applying one function to the result of another function. If we have two functions, f(x) and g(x), then fg(x) means apply function g first, and then apply function f to the result. This can also be written as f(g(x)). Note that fg(x) is generally not the same as gf(x).

fg(x) = f(g(x))

Inverse Functions

An inverse function, denoted f⁻¹(x), reverses the effect of the original function f(x). If f(a) = b, then f⁻¹(b) = a. When a function is composed with its inverse, the result is the original input, i.e., f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. To find the inverse function algebraically, follow these steps: 1. Let y = f(x). 2. Swap x and y. 3. Rearrange the equation to make y the subject. 4. Replace y with f⁻¹(x).

f(f⁻¹(x)) = x

Equation of a Circle (Centred at the Origin)

The equation of a circle with its centre at the origin (0, 0) and a radius 'r' can be derived using Pythagoras' theorem. For any point (x, y) on the circle, the distance from the origin to that point is 'r'. Forming a right-angled triangle with vertices at (0,0), (x,0), and (x,y), the lengths of the sides are x, y, and r (the hypotenuse). Thus, x² + y² = r².

x² + y² = r²

Key facts to remember

1f(x) represents the output of a function when the input is x.
2fg(x) means apply function g first, then function f to the result.
3f⁻¹(x) is the inverse function of f(x), which reverses the operation of f(x).
4To find f⁻¹(x), let y = f(x), swap x and y, then rearrange to make y the subject.
5The equation of a circle centred at the origin (0, 0) is x² + y² = r², where r is the radius.
6The radius 'r' must always be a positive value.
7To check if a point (x, y) lies on a circle, substitute its coordinates into the circle's equation. If the equation holds true, the point is on the circle.

Worked examples

Example 1

Given the functions f(x) = 3x - 2 and g(x) = x² + 1, find: (a) f(5) (b) fg(x) (c) gf(x)

I(a) To find f(5), substitute x=5 into f(x): f(5) = 3(5) - 2

IIf(5) = 15 - 2

IIIf(5) = 13

IV(b) To find fg(x), substitute g(x) into f(x): fg(x) = f(g(x)) = f(x² + 1)

VNow substitute (x² + 1) into the expression for f(x): fg(x) = 3(x² + 1) - 2

VIfg(x) = 3x² + 3 - 2

VIIfg(x) = 3x² + 1

VIII(c) To find gf(x), substitute f(x) into g(x): gf(x) = g(f(x)) = g(3x - 2)

9Now substitute (3x - 2) into the expression for g(x): gf(x) = (3x - 2)² + 1

10Expand the bracket: gf(x) = (3x - 2)(3x - 2) + 1

11gf(x) = 9x² - 6x - 6x + 4 + 1

12gf(x) = 9x² - 12x + 5

Answer

(a) f(5) = 13 (b) fg(x) = 3x² + 1 (c) gf(x) = 9x² - 12x + 5

Remember that fg(x) means 'g first, then f'.

Example 2

Given the function h(x) = (x + 4) / 5, find: (a) h⁻¹(x) (b) h⁻¹(3)

I(a) To find the inverse function h⁻¹(x): Step 1: Let y = h(x) y = (x + 4) / 5

IIStep 2: Swap x and y x = (y + 4) / 5

IIIStep 3: Rearrange to make y the subject 5x = y + 4

IV5x - 4 = y

VStep 4: Replace y with h⁻¹(x) h⁻¹(x) = 5x - 4

VI(b) To find h⁻¹(3), substitute x=3 into h⁻¹(x): h⁻¹(3) = 5(3) - 4

VIIh⁻¹(3) = 15 - 4

VIIIh⁻¹(3) = 11

Answer

(a) h⁻¹(x) = 5x - 4 (b) h⁻¹(3) = 11

Always show the steps for finding the inverse function, especially swapping x and y.

Example 3

A circle has the equation x² + y² = 36. (a) State the radius of the circle. (b) Does the point (4, 5) lie on the circle? Show your working. (c) Does the point (-6, 0) lie on the circle? Show your working.

I(a) The general equation of a circle centred at the origin is x² + y² = r². Comparing this to x² + y² = 36, we have r² = 36.

IITo find the radius r, take the square root of r²: r = √36

IIIr = 6 (Radius must be positive)

IV(b) To check if the point (4, 5) lies on the circle, substitute x=4 and y=5 into the equation x² + y² = 36: (4)² + (5)² = 36

V16 + 25 = 36

VI41 = 36

VIISince 41 ≠ 36, the point (4, 5) does not lie on the circle.

VIII(c) To check if the point (-6, 0) lies on the circle, substitute x=-6 and y=0 into the equation x² + y² = 36: (-6)² + (0)² = 36

936 + 0 = 36

1036 = 36

11Since 36 = 36, the point (-6, 0) lies on the circle.

Answer

(a) The radius of the circle is 6. (b) No, the point (4, 5) does not lie on the circle. (c) Yes, the point (-6, 0) lies on the circle.

Remember that r is the radius, so you need to square root the number on the right side of the equation.

Common mistakes

✗Confusing the order of composite functions, e.g., calculating gf(x) instead of fg(x).
✗Making algebraic errors when rearranging to find the inverse function, particularly with signs or fractions.
✗Forgetting to square root the constant term in the circle equation (r²) to find the radius (r).
✗Incorrectly substituting negative coordinates into the circle equation, e.g., (-3)² becoming -9 instead of 9.
✗Assuming that fg(x) is always equal to gf(x); they are generally different.

Exam tips

★For composite functions, always work from the inside out: fg(x) means substitute x into g, then substitute the result into f.
★When finding inverse functions, clearly show the step where you swap x and y, as this is a key part of the method.
★Always state the radius of a circle as a positive value. If r² is given, remember to take the positive square root.
★When checking if a point lies on a circle, substitute the coordinates carefully and show your calculation to justify your conclusion.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.

Solve a problem →← Maths curriculum