The equation of a circle with centre (h, k) and radius r is (x − h)² + (y − k)² = r² (standard form). Expanded, it becomes the general form x² + y² + Dx + Ey + F = 0, whose centre is (−D/2, −E/2) and radius is √((D/2)² + (E/2)² − F).
(x − h)² + (y − k)² = r² — centre (h, k), radius rx² + y² + Dx + Ey + F = 0(−D/2, −E/2)r = √((D/2)² + (E/2)² − F) = ½√(D² + E² − 4F)(x₁, y₁), (x₂, y₂): (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0= r (less than r: 2 intersection points; greater: none)Expanding (x − h)² + (y − k)² = r² gives the general form x² + y² + Dx + Ey + F = 0, where D = −2h, E = −2k and F = h² + k² − r².
Three checks tell you an equation is a circle: the x² and y² terms have the same coefficient, there is no xy term, and after dividing by that coefficient D² + E² − 4F > 0. If D² + E² − 4F = 0 the equation is a single point; if it is negative there is no real circle.
From the standard form, read them off: centre (h, k), radius r. Watch the signs: (x + 3)² + (y − 1)² = 25 has centre (−3, 1) and radius 5.
From the general form:
= (−D/2, −E/2).= √((D/2)² + (E/2)² − F).You can also complete the square on x and y to rewrite the equation in standard form. Examples 1 and 2 below show both methods.
(h, k)). In the DSE, you must be comfortable switching between Standard Form (which tells you the center and radius immediately) and General Form (which requires algebraic formulas or completing the square).Scan this compact reference table to review every essential concept, step, and shortcut in under 10 seconds.
| Topic | Key Concept & Steps | Exam Shortcuts & Tips |
|---|---|---|
| 1. Standard Form | (x − h)² + (y − k)² = r²• Centre: (h, k)• Radius: r | Tip: Pay attention to the signs.(x + 3)² ⟹ h = −3. |
| 2. General Form | x² + y² + Dx + Ey + F = 0• Centre: (−D/2, −E/2)• Radius: r = √(D²/4 + E²/4 − F) | Real circle condition:D² + E² − 4F > 0. If ≤ 0, no real circle exists. |
| 3. Diameter Form | Given endpoints A(x₁, y₁) and B(x₂, y₂):(x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0 | A massive time-saver for Paper 2. You do not need to find the midpoint or radius to write this equation. |
| 4. Point vs. Circle | Compare distance d from point P to centre C with radius r: • d > r: Outside the circle • d = r: On the circle • d < r: Inside the circle | Alternatively, substitute P(x₁, y₁) into the general equation:• Result > 0: Outside • Result = 0: On • Result < 0: Inside |
| 5. Line Intersections | Substitute y = mx + c into the circle equation. Solve the resulting quadratic using the discriminant (Δ = b² − 4ac).• Δ > 0: 2 intersection points • Δ = 0: 1 intersection point (Tangent) • Δ < 0: 0 intersection points | For tangents, use the geometric shortcut: perpendicular distance from centre to line = radius (d = r). Much faster than solving Δ = 0. |
Find the centre and radius of the circle x² + y² − 6x + 8y − 11 = 0.
D = −6, E = 8, F = −11.= (−D/2, −E/2) = (3, −4).= √(3² + (−4)² − (−11)) = √(9 + 16 + 11) = √36 = 6.(x − 3)² + (y + 4)² = 11 + 9 + 16 = 36 ✓Answer: Centre (3, −4), radius 6.
Write 2x² + 2y² + 8x − 12y − 6 = 0 in standard form.
x² + y² + 4x − 6y − 3 = 0.(x² + 4x + 4) + (y² − 6y + 9) = 3 + 4 + 9.(x + 2)² + (y − 3)² = 16.Answer: (x + 2)² + (y − 3)² = 16: centre (−2, 3), radius 4.
Find the equation of the circle with diameter AB, where A(1, 2) and B(5, −4).
(x − 1)(x − 5) + (y − 2)(y + 4) = 0.x² − 6x + 5 + y² + 2y − 8 = 0.x² + y² − 6x + 2y − 3 = 0.(3, −1) is the mid-point of AB, and r = √(9 + 1 + 3) = √13 is half of AB = √52 ✓Answer: x² + y² − 6x + 2y − 3 = 0
Show that the line 2x + y − 5 = 0 is a tangent to the circle x² + y² = 5, and find the point of contact.
(0, 0), radius √5.= |2(0) + 0 − 5| / √(2² + 1²) = 5/√5 = √5, which equals the radius, so the line is a tangent.y = 5 − 2x to get x² + (5 − 2x)² = 5, i.e. 5x² − 20x + 20 = 0, so (x − 2)² = 0.x = 2 and y = 5 − 4 = 1.Answer: The line is a tangent; the point of contact is (2, 1).
3x² + 3y² − 12x + 18y − 9 = 0, you must divide the entire equation by 3 first to get x² + y² − 4x + 6y − 3 = 0. Only then can you state D = −4, E = 6, F = −3.D² + E² − 4F > 0. If D² + E² − 4F = 0, the equation represents a single point. If it is < 0, no real circle exists.y = mx + c and solving Δ = 0 is very slow. Shortcut: Calculate the perpendicular distance d from the circle's center to the line using the point-to-line formula. Set d = r (distance equals radius). This is usually twice as fast as the algebraic Δ method.How to study: Click any question below to instantly load it on the coordinate grid and check your steps.
x² + y² − 6x + 8y − 11 = 0.y = 2x + 1 and the circle x² + y² = 25.(5, −2) lies inside, on, or outside the circle (x − 2)² + (y + 1)² = 16.DSE-style practice sets from the SketchMath question bank, with solutions for every MC and long question.