The equation of a circle · Grade 10 · 5 units
In The equation of a circle (Grade 10 · 5 units) 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
For example: What is the radius of the circle x² + y² = 4 ?
Answer and solution
The answer: 2
Here r² = 4 , so r = √4 = 2 .
The centre and radius from the equation
For example: What is the centre of the circle (x + 5)² + (y − 6)² = 1 ?
Answer and solution
The answer: (−5, 6)
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
For example: What is the equation of the circle with centre M(−5, 6) and radius 2?
- (x − 5)² + (y + 6)² = 4
- (x − 5)² + (y − 6)² = 4
- (x + 5)² + (y − 6)² = 2
- (x + 5)² + (y − 6)² = 4
Answer and solution
The answer: open bracket x plus five close bracket squared plus open bracket y minus six close bracket squared equals four
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
For example: Is the point P(−5, 6) on the circle x² + y² = 61 , inside it or outside it?
- Outside the circle
- Inside the circle
- On the circle
Answer and solution
The answer: On the circle
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
For example:
At which points does the line x = 4 meet the circle x² + y² = 25 ?- (−4, −3), (−4, 3)
- (4, −3), (4, 3)
- (−4, −3), (4, 3)
- (−3, 4), (3, 4)
Answer and solution
The answer: open bracket four, minus three close bracket, open bracket four, three close bracket
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
For example:
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 and solution
The answer: 5
The vertex is on the circle, so the radius is its distance from the origin: r = √((−3)² + 4²) = √25 = 5 .
Not on the site yet
Parts of this topic that are not built yet: The general equation of a circle (not in the programme) · A tangent to a circle in coordinate geometry.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).