01 · Write an equation
Find the circle with centre (−2,5) and radius √13.
Hint
Use (x − a)² + (y − b)² = r².
Worked solution
(x + 2)² + (y − 5)² = 13
Understand · explore · practise
Equations of circles
Find a circle equation from its centre, radius or diameter. Complete the square to recover the centre and radius, and test whether a point is inside, on or outside a circle.
Before you startPythagoras, midpoint formula and completing the square
01 / Circle equation
Every point on the circle has the same distance from its centre.
A circle with centre C(a,b) and positive radius r consists of the points (x,y) whose distance from C is r. Pythagoras gives:
(x − a)² + (y − b)² = r²
The horizontal and vertical changes are x − a and y − b. At the origin this simplifies to x² + y² = r². Changing the centre translates the circle.
In the model, change one coordinate at a time. The sign inside each bracket is opposite to the corresponding centre coordinate.
Change the centre and radiusEqual axis scales
Centre C(2,−2), radius 3. The equation is (x − 2)² + (y + 2)² = 9. Both axes have the same unit scale.
Watch Pythagoras produce the circle equation
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Read the equation
The right-hand side is r², not r.
For (x + 3)² + (y − 2)² = 20, the centre is (−3,2) and radius is √20 = 2√5.
Centre (4, −1), radius 3√2
⇒ (x − 4)² + (y + 1)² = 18
Do not confuse the circle with its filled interior: equality describes the circumference. Replacing = by < describes the points strictly inside.
A parameter in the radius
(x − 2a)² + (y + a)² = 9a² has centre (2a,−a) and radius 3|a| when a ≠ 0. For a = 0 it describes one point. Writing the radius as 3a would require a > 0.
03 / Diameter
Find the centre and radius squared from the endpoints.
The centre is the midpoint of a diameter. The radius is half its length, so its square is one quarter of the diameter’s squared length.
r² = [(x₂ − x₁)² + (y₂ − y₁)²]/4
This often avoids taking a square root only to square it again. You can also find r² using the distance from the midpoint to either endpoint.
A diameter determines a circleWorked example
A = (−4, 1), B = (6, 5)
The midpoint is C(1,3).
AB² = 10² + 4² = 116
r² = 116/4 = 29
Use one quarter, not one half, of AB².
(x − 1)² + (y − 3)² = 29
Both endpoints satisfy the equation.
04 / Centre from a chord
A known chord puts the centre on its perpendicular bisector.
Two points on a circle usually leave infinitely many possible centres along their perpendicular bisector. Extra information, such as one centre coordinate or the radius, can narrow the possibilities.
If A(−3,0), B(3,0) are on a circle of radius 5, the centre lies on x = 0. Its distance from the chord is √(25 − 9) = 4, giving centres (0,4) and (0,−4). A requirement that the centre is above the axis selects the first.
Use a known centre coordinateWorked example
P = (−2,1), Q = (4,3)
The centre has y-coordinate −1
The chord midpoint is (1,2), with chord gradient 1/3.
Bisector: y − 2 = −3(x − 1)
y = −3x + 5
The centre lies on this line.
−1 = −3x + 5 ⇒ x = 2
C = (2,−1)
Use either endpoint to calculate the radius.
r² = (−2 − 2)² + (1 + 1)² = 20
(x − 2)² + (y + 1)² = 20
The other endpoint gives the same squared radius.
05 / Test a point
Compare its squared distance with the radius squared.
For centre C and point P, compare CP² with r². Smaller means inside, equal means on the circle, and larger means outside.
Circle: (x − 2)² + (y + 1)² = 25
P(5,3): CP² = 3² + 4² = 25 → on
Q(2,2): CQ² = 0² + 3² = 9 → inside
R(8,−1): CR² = 6² = 36 → outside
If you know the centre and any point on the circle, their squared distance supplies r². One arbitrary point alone does not determine a unique circle.
06 / Complete squares
Complete the square once for x and once for y.
Group the x terms and the y terms before completing the square. Keep track of both correction constants.
Recover the geometry from expanded formWorked example
x² + y² − 6x + 8y − 11 = 0
Group x² − 6x and y² + 8y.
(x − 3)² − 9
+ (y + 4)² − 16 − 11 = 0
Both added squares need compensating constants.
(x − 3)² + (y + 4)² = 36
Move the constants to the right.
Centre (3, −4), radius 6
Expand the brackets to check the original equation.
07 / Is it a circle?
Check the coefficients and the resulting radius squared.
For x² + y² + 2fx + 2gy + c = 0, completing squares gives:
(x + f)² + (y + g)² = f² + g² − c
The centre is (−f,−g). The right side must be positive for a circle of positive radius. Zero gives a single point; a negative value gives no real points.
If x² and y² have the same non-zero coefficient, first divide the entire equation by it. There must be no xy term. Unequal squared coefficients, such as 4x² + y² = 16, do not describe a circle in ordinary equal-scale coordinates.
2x² + 2y² − 8x + 4y − 6 = 0
⇒ (x − 2)² + (y + 1)² = 8
Here the centre is (2,−1) and radius is 2√2.
08 / Parameters
Use a point to find an unknown centre or radius.
Substitution can give two possible circles. Keep both unless the question supplies an extra condition.
(x − k)² + (y + 1)² = 25
passes through (2,3)
(2 − k)² + 16 = 25
k = −1 or 5
For x² + y² + 4x − 6y = k, completing squares gives (x + 2)² + (y − 3)² = k + 13. A circle of positive radius therefore requires k > −13.
09 / Your turn
Check the signs of the centre and the square of the radius.
A quick substitution often catches a sign error.
02 · Read the geometry
(x − 1)² + (y + 4)² = 45
Hint
Take the positive square root of 45.
Worked solution
Centre (1,−4), radius 3√5.
03 · Centre and one point
The centre is (3,−2) and the circle passes through (−1,1). Find its equation.
Hint
The displacement is (−4,3).
Worked solution
r² = 16 + 9 = 25
(x − 3)² + (y + 2)² = 25
04 · Diameter endpoints
Find the circle with diameter endpoints (−2,−1) and (4,7).
Hint
Centre (1,3); diameter length 10.
Worked solution
(x − 1)² + (y − 3)² = 25
05 · Expanded form
Find the centre and radius of x² + y² + 10x − 2y + 1 = 0.
Hint
Complete both squares.
Worked solution
(x + 5)² + (y − 1)² = 25
Centre (−5,1), radius 5.
06 · Divide first
Find the centre and radius of 3x² + 3y² − 6x + 12y − 12 = 0.
Hint
Divide every term by 3.
Worked solution
x² + y² − 2x + 4y − 4 = 0
(x − 1)² + (y + 2)² = 9
Centre (1,−2), radius 3.
07 · Two possible centres
(x − k)² + (y − 2)² = 10 passes through (1,5). Find k.
Hint
(1 − k)² + 9 = 10.
Worked solution
(1 − k)² = 1
k = 0 or 2
08 · Circle, point or nothing?
Classify x² + y² − 2x + 6y = k according to k.
Hint
(x − 1)² + (y + 3)² = k + 10.
Worked solution
k > −10: circle. k = −10: the single point (1,−3). k < −10: no real points.
09 · A centre coordinate
A circle through (−1,0) and (5,2) has centre y-coordinate −2. Find its equation.
Hint
The chord midpoint is (2,1), and its perpendicular bisector is y = −3x + 7.
Worked solution
−2 = −3x + 7 ⇒ C = (3,−2)
r² = (−1 − 3)² + (0 + 2)² = 20
(x − 3)² + (y + 2)² = 20
10 / Recap
A circle equation is a distance statement.
- Read the centre with the correct bracket signs.
- The right side is r².
- A diameter gives the midpoint and one quarter of its squared length.
- Complete two squares in expanded equations.
- Check r² > 0 before claiming a circle.
- Compare squared distances to classify points.
Section 1 of 10 · Circle equation