# The equation of a circle · Years 11–12

> The equation of a circle, Years 11–12: A point and a circle and more. Explanation, worked examples and free practice for the national curriculum in England.

In The equation of a circle (Years 11–12) you practise 6 skills: The circle centred at the origin · The centre and radius from the equation · Writing the equation of a circle · A point and a circle · A circle and a line · A square inscribed in a circle.

A circle is every point at a fixed distance, the radius, from one point, the centre. By the distance formula, a point is on the circle when the squares of the differences between its coordinates and the centre’s add up to the radius squared. With the centre at the origin, the equation is x squared plus y squared equals the radius squared.

With the centre somewhere else, each bracket has a minus before the centre’s coordinate, so a plus in the bracket means the centre’s coordinate is negative. The right side is the radius squared: take its square root to find the radius. Putting a point into the left side gives its squared distance from the centre, which tells you whether it is on the circle, inside it or outside it.

The points where a circle and a line meet satisfy both equations. Put the line’s equation into the circle’s to get one quadratic equation. For each solution, the line gives the other coordinate.

**Tip:** Before you put anything in, write down the centre and the radius squared, and take care with the signs inside the brackets.

## What you learn

### The circle centred at the origin

- Example: What is the radius of the circle x² + y² = 4 ?
- Answer: 2
- Solution: Here r² = 4 , so r = √4 = 2 .

### The centre and radius from the equation

- Example: What is the centre of the circle (x + 5)² + (y − 6)² = 1 ?
- Answer: (−5, 6)
- Solution: Compare with (x − a)² + (y − b)² = r² : x + 5 = x − (−5) means a = −5 , and y − 6 means b = 6 . The centre is M(−5, 6) .

### Writing the equation of a circle

- Example: What is the equation of the circle with centre M(−5, 6) and radius 2?
- Options: open bracket x minus five close bracket squared plus open bracket y plus six close bracket squared equals four (x − 5)² + (y + 6)² = 4 · open bracket x minus five close bracket squared plus open bracket y minus six close bracket squared equals four (x − 5)² + (y − 6)² = 4 · open bracket x plus five close bracket squared plus open bracket y minus six close bracket squared equals two (x + 5)² + (y − 6)² = 2 · open bracket x plus five close bracket squared plus open bracket y minus six close bracket squared equals four (x + 5)² + (y − 6)² = 4
- Answer: open bracket x plus five close bracket squared plus open bracket y minus six close bracket squared equals four
- Solution: Put a = −5, b = 6, r = 2 into (x − a)² + (y − b)² = r² to get (x + 5)² + (y − 6)² = 4 .

### A point and a circle

- Example: Is the point P(−5, 6) on the circle x² + y² = 61 , inside it or outside it?
- Options: Outside the circle · Inside the circle · On the circle
- Answer: On the circle
- Solution: Put the point into the left side: (−5)² + 6² = 25 + 36 = 61 . That equals the right side, so the point is on the circle.

### A circle and a line

- Example: (A circle and a line cutting it at two points.) At which points does the line x = 4 meet the circle x² + y² = 25 ?
- Options: open bracket minus four, minus three close bracket, open bracket minus four, three close bracket (−4, −3), (−4, 3) · open bracket four, minus three close bracket, open bracket four, three close bracket (4, −3), (4, 3) · open bracket minus four, minus three close bracket, open bracket four, three close bracket (−4, −3), (4, 3) · open bracket minus three, four close bracket, open bracket three, four close bracket (−3, 4), (3, 4)
- Answer: open bracket four, minus three close bracket, open bracket four, three close bracket
- Solution: Put x = 4 into the circle’s equation: 4² + y² = 25 , that is y² = 9 , so y = 3 or y = −3 . The points are (4, −3), (4, 3) .

### A square inscribed in a circle

- Example: (A circle centred at the origin, with a vertex of a square on it.) The square ABCD is inscribed in a circle centred at the origin, and one of its vertices is A(−3, 4) . What is the radius of the circle?
- Answer: 5
- Solution: The vertex is on the circle, so the radius is its distance from the origin: r = √((−3)² + 4²) = √25 = 5 .

The exercises are made by rules and the answers are worked out in code. The topic follows the national curriculum in England: mathematics (key stages 3 and 4) and selected AS and A level mathematics practice in England (the GCE subject content).

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