What is a circle in coordinates?
A circle is the set of all points that are the same distance from one fixed point. The fixed point is the centre. The fixed distance is the radius r.
Take the centre C(h, k) and any point P(x, y) on the circle. The distance formula says CP = √((x − h)² + (y − k)²). This must equal r. Square both sides:
(x − h)² + (y − k)² = r² (centre–radius form, also called standard form)
If the centre is the origin, h = k = 0 and the equation is x² + y² = r².
A point is on the circle if it makes the equation true. If the left side is smaller than r², the point is inside; if larger, it is outside.
General form: finding the centre and radius
Open the brackets of (x − h)² + (y − k)² = r²:
x² + y² − 2hx − 2ky + (h² + k² − r²) = 0.
So every circle can be written as x² + y² + Dx + Ey + F = 0. Some books write it as x² + y² + 2gx + 2fy + c = 0.
- Centre = (−D/2, −E/2), or (−g, −f).
- r² = (D/2)² + (E/2)² − F, or r² = g² + f² − c.
To go back, complete the square in x and in y. Example: x² + y² − 6x + 4y − 3 = 0 → (x − 3)² − 9 + (y + 2)² − 4 − 3 = 0 → (x − 3)² + (y + 2)² = 16. Centre (3, −2), radius 4.
When is it really a circle?
The x² and y² terms must have the same coefficient and there must be no xy term. If r² comes out positive, it is a real circle. If r² = 0 it is just one point. If r² is negative, no point fits.
A line and a circle
Find d, the perpendicular distance from the centre (h, k) to the line ax + by + c = 0: d = |ah + bk + c| / √(a² + b²).
- d < r: the line cuts the circle in 2 points (a secant).
- d = r: the line touches at 1 point (a tangent). The radius to that point is perpendicular to the line.
- d > r: no common point.
The algebra way
Put y from the line into the circle. You get a quadratic in x. Its discriminant Δ = b² − 4ac tells the same story: Δ > 0 two points, Δ = 0 one point, Δ < 0 none. Solving the quadratic gives the actual meeting points.
Tangent at a point on the circle
For x² + y² = r², the tangent at (x₁, y₁) is x₁x + y₁y = r². In general, it is the line through (x₁, y₁) perpendicular to the radius.
Two circles
Let the radii be r₁ and r₂ and the distance between the centres be d.
- d > r₁ + r₂: apart, no common point.
- d = r₁ + r₂: they touch outside, 1 point.
- |r₁ − r₂| < d < r₁ + r₂: they cut in 2 points.
- d = |r₁ − r₂|: they touch inside, 1 point.
- d < |r₁ − r₂|: one is inside the other, no point (d = 0 means same centre: concentric).
To find the meeting points, subtract one equation from the other. The x² and y² cancel and leave a straight line (the common chord). Solve this line with either circle.
Try it
Draw a radius-5 circle about (0, 0) on squared paper with a pin and thread. Find all 12 grid corners on it: (±3, ±4), (±4, ±3), (±5, 0), (0, ±5). In the 3D, choose step 4 and slide the line: watch the yellow meeting points go from 2 to 1 to 0 as d passes r.
Key formulas and definitions
- Centre (h, k), radius r: (x − h)² + (y − k)² = r²
- Centre at origin: x² + y² = r²
- General form: x² + y² + Dx + Ey + F = 0, centre (−D/2, −E/2), r² = D²/4 + E²/4 − F
- Distance from (h, k) to ax + by + c = 0: d = |ah + bk + c| / √(a² + b²)
- Line vs circle: d < r → 2 points; d = r → tangent; d > r → none
- Two circles: compare d with r₁ + r₂ and |r₁ − r₂|
- Tangent to x² + y² = r² at (x₁, y₁): x₁x + y₁y = r²
Worked examples
1. Write the equation of the circle with centre (0, 0) and radius 6.
x² + y² = 6² → x² + y² = 36.
2. Write the equation of the circle with centre (2, −3) and radius 4.
(x − 2)² + (y − (−3))² = 4² → (x − 2)² + (y + 3)² = 16.
3. Find the centre and radius of (x + 1)² + (y − 5)² = 49.
Flip the signs: centre (−1, 5). r² = 49, so r = 7.
4. Find the centre and radius of x² + y² − 8x + 2y + 8 = 0.
Group: (x² − 8x) + (y² + 2y) = −8. Complete squares: (x − 4)² − 16 + (y + 1)² − 1 = −8 → (x − 4)² + (y + 1)² = 9. Centre (4, −1), radius 3.
5. Find the circle with endpoints of a diameter A(1, 2) and B(7, 10).
Centre = midpoint = ((1 + 7)/2, (2 + 10)/2) = (4, 6). r = distance from (4, 6) to (1, 2) = √(9 + 16) = 5. Equation: (x − 4)² + (y − 6)² = 25.
6. Does the line y = x + 1 meet the circle x² + y² = 5? Find the points.
Substitute: x² + (x + 1)² = 5 → 2x² + 2x − 4 = 0 → x² + x − 2 = 0 → (x + 2)(x − 1) = 0. x = −2 gives y = −1; x = 1 gives y = 2. Two points: (−2, −1) and (1, 2). Check with d: d = |0 − 0 + 1|/√2 ≈ 0.71 < √5 ≈ 2.24.
7. For which value of c is the line 3x + 4y + c = 0 a tangent to x² + y² = 4 (c > 0)?
Tangent means d = r. d = |c| / √(9 + 16) = |c|/5. Set |c|/5 = 2 → c = 10.
8. Circles x² + y² = 9 and (x − 5)² + y² = 4: how are they placed?
Centres (0, 0) and (5, 0), so d = 5. r₁ = 3, r₂ = 2, r₁ + r₂ = 5 = d. They touch outside at one point, (3, 0).
Common mistakes
- Keeping the sign: (x − 2)² + (y + 3)² = 16 has centre (2, −3), not (−2, 3).
- Forgetting the square: the right side is r², so r² = 16 means r = 4, not 16.
- Calling any second-degree equation a circle: x² + 2y² = 4 is not a circle because the coefficients of x² and y² differ.
- Comparing d with the diameter instead of the radius when testing a line against a circle.