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9.2 Standard and General Equations of Circles

Key Takeaways

  • The standard equation (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2 (r>0r > 0) reveals center C(a,b)C(a, b) and radius rr, while general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 represents a real circle if and only if D2+E2−4F>0D^2 + E^2 - 4F > 0.

  • Parametric equations x=a+rcos⁡θ,y=b+rsin⁡θx = a + r\cos\theta, y = b + r\sin\theta (θ∈[0,2π)\theta \in [0, 2\pi)) convert distance and optimization problems on circles into single-variable trigonometric functions.

  • An Apollonius circle is the locus of a point PP moving such that the ratio of its distances to two fixed points AA and BB is a constant k≠1k \neq 1 (∣PA∣=k∣PB∣|PA| = k |PB|).

  • Positional relationships of two circles depend on center distance dd vs radii R,rR, r; subtracting intersecting circle equations yields their radical axis and common chord of length L=2R2−dcenter2L = 2\sqrt{R^2 - d_{center}^2}.

Last updated: July 2026

9.2 Standard and General Equations of Circles

A circle is the geometric locus of all points in a plane that are at a fixed distance (the radius r>0r > 0) from a fixed point (the center C(a,b)C(a, b)). In Gaokao analytic geometry, questions on circles test structural algebraic transformations, geometric properties (such as chord length, tangency, and symmetry), parametric optimizations, locus derivations, and interactions between multiple circles.


1. Standard and General Forms of Circle Equations

Standard Form

Applying the distance formula between a dynamic point P(x,y)P(x, y) and fixed center C(a,b)C(a, b) set equal to radius r>0r > 0:

(x−a)2+(y−b)2=r  ⟹  (x−a)2+(y−b)2=r2\sqrt{(x - a)^2 + (y - b)^2} = r \implies (x - a)^2 + (y - b)^2 = r^2
  • Key Advantage: Standard Form explicitly reveals the center coordinates (a,b)(a, b) and radius rr. It is preferred when geometric conditions involve center coordinates or radius explicitly.

General Form

Expanding the standard form equation:

x2−2ax+a2+y2−2by+b2−r2=0x^2 - 2ax + a^2 + y^2 - 2by + b^2 - r^2 = 0

Letting D=−2aD = -2a, E=−2bE = -2b, and F=a2+b2−r2F = a^2 + b^2 - r^2, we obtain the General Form:

x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0

To recover geometric parameters from the general form, complete the square for xx and yy:

(x+D2)2+(y+E2)2=D2+E2−4F4\left(x + \frac{D}{2}\right)^2 + \left(y + \frac{E}{2}\right)^2 = \frac{D^2 + E^2 - 4F}{4}

Discriminant Analysis of General Form

The equation x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 is governed by the discriminant Δcircle=D2+E2−4F\Delta_{circle} = D^2 + E^2 - 4F:

  1. Δcircle>0\Delta_{circle} > 0: Represents a real circle with center C(−D2,−E2)C\left(-\frac{D}{2}, -\frac{E}{2}\right) and radius r=12D2+E2−4Fr = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}.
  2. Δcircle=0\Delta_{circle} = 0: Represents a single point C(−D2,−E2)C\left(-\frac{D}{2}, -\frac{E}{2}\right) (a degenerate circle with radius r=0r = 0).
  3. Δcircle<0\Delta_{circle} < 0: Represents no real locus (empty set in R2\mathbb{R}^2).

Gaokao Caution: Never assume an equation of the form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 represents a circle without first checking or enforcing D2+E2−4F>0D^2 + E^2 - 4F > 0.


2. Parametric Equations of a Circle

For a circle (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2, introducing parameter angle θ∈[0,2π)\theta \in [0, 2\pi) (measured counterclockwise from the positive horizontal direction centered at C(a,b)C(a, b)) gives the parametric equations:

{x=a+rcos⁡θy=b+rsin⁡θ(θ∈[0,2π))\begin{cases} x = a + r \cos \theta \\ y = b + r \sin \theta \end{cases} \quad (\theta \in [0, 2\pi))

Applications of Parametric Equations

  • Algebraic Extremum Problems: To find the maximum or minimum of a linear objective function z=Ax+Byz = Ax + By for points (x,y)(x, y) on circle (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2, substitute parametric coordinates: z=A(a+rcos⁡θ)+B(b+rsin⁡θ)=(Aa+Bb)+r(Acos⁡θ+Bsin⁡θ)=(Aa+Bb)+rA2+B2sin⁡(θ+ϕ)z = A(a + r\cos\theta) + B(b + r\sin\theta) = (Aa + Bb) + r(A\cos\theta + B\sin\theta) = (Aa + Bb) + r\sqrt{A^2 + B^2}\sin(\theta + \phi) where tan⁡ϕ=AB\tan\phi = \frac{A}{B}. Thus zmax⁡=Aa+Bb+rA2+B2z_{\max} = Aa + Bb + r\sqrt{A^2 + B^2} and zmin⁡=Aa+Bb−rA2+B2z_{\min} = Aa + Bb - r\sqrt{A^2 + B^2}.

3. Methods for Determining Equations of Circles

Determining a circle's equation requires finding three independent algebraic coefficients (either a,b,ra, b, r or D,E,FD, E, F). Two principal methods exist:

Algebraic Method (Undetermined Coefficients)

  1. Assume general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0.
  2. Set up a system of three linear equations using three given non-collinear points P1,P2,P3P_1, P_2, P_3.
  3. Solve for D,E,FD, E, F and verify D2+E2−4F>0D^2 + E^2 - 4F > 0.

Geometric Method

  1. Use geometric properties to locate center C(a,b)C(a, b):
    • Center CC lies on the perpendicular bisectors of any chords formed by given points.
    • Center CC lies at distance rr from tangent lines.
  2. Compute radius r=∣CP1∣r = |CP_1|.
  3. Write the standard form (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2.

Diameter Form of a Circle

If segment ABAB with endpoints A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) is a diameter of circle CC, then any point P(x,y)P(x, y) on the circle (except A,BA, B) satisfies PA⃗⊥PB⃗  ⟺  PA⃗⋅PB⃗=0\vec{PA} \perp \vec{PB} \iff \vec{PA} \cdot \vec{PB} = 0. Expanded equation:

(x−x1)(x−x2)+(y−y1)(y−y2)=0(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0

Apollonius Circle (阿波罗尼斯圆)

In the plane, the locus of point PP such that the ratio of its distances to two fixed points AA and BB is a constant kk (k>0,k≠1k > 0, k \neq 1) is a circle:

∣PA∣∣PB∣=k  ⟹  ∣PA∣2=k2∣PB∣2\frac{|PA|}{|PB|} = k \implies |PA|^2 = k^2 |PB|^2

Derivation: Place A(−c,0)A(-c, 0) and B(c,0)B(c, 0) on the xx-axis. Expanding (x+c)2+y2=k2[(x−c)2+y2](x + c)^2 + y^2 = k^2 [(x - c)^2 + y^2] yields a quadratic equation with equal non-zero coefficients for x2x^2 and y2y^2 and no xyxy term, representing a circle.


4. Positional Relationships Between Two Circles

Let two circles be C1:(x−a1)2+(y−b1)2=R2C_1: (x - a_1)^2 + (y - b_1)^2 = R^2 and C2:(x−a2)2+(y−b2)2=r2C_2: (x - a_2)^2 + (y - b_2)^2 = r^2 with R≥r>0R \ge r > 0. Let d=∣C1C2∣=(a1−a2)2+(b1−b2)2d = |C_1 C_2| = \sqrt{(a_1 - a_2)^2 + (b_1 - b_2)^2} be the center-to-center distance.

Positional RelationshipDistance CriterionNumber of Intersection PointsNumber of Common Tangents
Outer Separation (外离)d>R+rd > R + r0044 (2 outer, 2 inner)
External Tangency (外切)d=R+rd = R + r1133 (2 outer, 1 inner)
Intersection (相交)∣R−r∣<d<R+r\vert R - r\vert< d < R + r2222 (2 outer)
Internal Tangency (内切)d=R−r>0d = R - r > 01111 (1 outer)
Inner Inclusion (内含)0≤d<R−r0 \le d < R - r0000

Common Chord and Radical Axis (相交弦与等幂轴)

When two circles C1:x2+y2+D1x+E1y+F1=0C_1: x^2 + y^2 + D_1 x + E_1 y + F_1 = 0 and C2:x2+y2+D2x+E2y+F2=0C_2: x^2 + y^2 + D_2 x + E_2 y + F_2 = 0 intersect at two points AA and BB, subtracting C2C_2 from C1C_1 eliminates x2x^2 and y2y^2:

(D1−D2)x+(E1−E2)y+(F1−F2)=0(D_1 - D_2)x + (E_1 - E_2)y + (F_1 - F_2) = 0

This linear equation represents the radical axis (等幂轴) of the two circles. For intersecting circles, it is precisely the line equation of their common chord ABAB.

  • Common Chord Length Formula: Perpendicular distance d1d_1 from center C1C_1 to line ABAB gives chord length: LAB=2R2−d12L_{AB} = 2 \sqrt{R^2 - d_1^2}
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Classification of Two-Circle Positional Relationships and Common Chord Derivation
Test Your Knowledge

For what value of parameter mm does the equation x2+y2−2x+4y+m=0x^2 + y^2 - 2x + 4y + m = 0 represent a circle with radius r=3r = 3?

A

m=5m = 5

B

m=0m = 0

C

m=−5m = -5

D

m=−4m = -4

Test Your Knowledge

What is the length of the common chord of the two intersecting circles C1:x2+y2=10C_1: x^2 + y^2 = 10 and C2:x2+y2−4x−4y+6=0C_2: x^2 + y^2 - 4x - 4y + 6 = 0?

A

222\sqrt{2}

B

10\sqrt{10}

C

252\sqrt{5}

D

2\sqrt{2}

Test Your Knowledge

Given two fixed points A(−1,0)A(-1, 0) and B(2,0)B(2, 0), a moving point P(x,y)P(x, y) satisfies ∣PA∣=2∣PB∣|PA| = 2 |PB|. What is the standard equation of the locus of PP (the Apollonius Circle)?

A

(x−2)2+y2=3(x - 2)^2 + y^2 = 3

B

(x−3)2+y2=9(x - 3)^2 + y^2 = 9

C

(x−3)2+y2=4(x - 3)^2 + y^2 = 4

D

(x−1)2+y2=4(x - 1)^2 + y^2 = 4

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