Understanding the Family of Circles in Geometry

A family of circles is a set of circles that share a common geometric condition, such as passing through a given point, intersecting at a fixed locus, or having centers restricted to a line. Such families are described by canonical forms in coordinate geometry and play a central role in the study of circles in two-dimensional Cartesian coordinates.

General Equation of a Circle in the Plane

In Cartesian coordinates, the general equation of a circle with center at $(h, k)$ and radius $r$ is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] On expanding, this becomes: \[ x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2 \] \[ x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0 \] Defining $2g = -2h$, $2f = -2k$ and $c = h^2 + k^2 - r^2$, the general form is: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] where $g$, $f$, $c$ are real constants.

Canonical Equations for Families of Circles Defined by a Geometric Condition

A family of circles is represented algebraically by introducing one or more parameters satisfying a locus or incidence property. These forms appear frequently in competitive examinations such as JEE Main. For comprehensive study of families of circles, see the Family Of Circles resource.

Family of Circles with Fixed Center

Let $(h, k)$ be a fixed point. The set of all circles centered at $(h, k)$ with varying radii forms a family, whose equation is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Here, the parameter $r$ varies across all non-negative real values. For each value of $r$, one obtains a specific circle. The family contains an infinite number of circles, all centered at $(h, k)$.

Family of Circles Passing Through a Fixed Point

Let $(x_1, y_1)$ be a fixed point. The set of all circles passing through $(x_1, y_1)$ can be described by the general equation of a circle, with the additional requirement that $(x_1, y_1)$ satisfies it: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] Substituting $x = x_1$, $y = y_1$: \[ x_1^2 + y_1^2 + 2g x_1 + 2f y_1 + c = 0 \] This imposes a linear relation among $(g, f, c)$, reducing the number of free parameters by one. The resulting family is parameterised by two real parameters, generally $g$ and $f$ or similar combinations.

Family of Circles Passing Through Two Fixed Points

Let $(x_1, y_1)$ and $(x_2, y_2)$ be two distinct fixed points. The family of all circles passing through both points is obtained as follows:

The general equation of a circle: \[ x^2 + y^2 + 2g x + 2f y + c = 0 \] Substitute $(x_1, y_1)$: \[ x_1^2 + y_1^2 + 2g x_1 + 2f y_1 + c = 0 \] Substitute $(x_2, y_2)$: \[ x_2^2 + y_2^2 + 2g x_2 + 2f y_2 + c = 0 \] These two linear equations in $g, f, c$ allow parameterisation by a single parameter. Solving for $g$ and $f$ in terms of $c$, and substituting back, gives a one-parameter family.

Alternatively, the equation of any circle with diameter $AB$, where $A(x_1, y_1)$ and $B(x_2, y_2)$, is: \[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \] This circle passes through $A$ and $B$ and is unique for given $A$ and $B$. The general family with arbitrary diameter between $A$ and $B$ is the set of all such equations as $A, B$ vary over a fixed locus.

Family of Circles Passing Through Intersection of Two Given Circles

Consider two circles \[ S_1 \equiv x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0 \] \[ S_2 \equiv x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0 \] Let $C$ be any point lying at the intersection $S_1 = 0,\, S_2 = 0$. Any circle passing through both common points has the following form: \[ S_1 + \lambda S_2 = 0;\quad \lambda \in \mathbb{R},\ \lambda \neq -1 \] To show this, note that for any real $\lambda$, $S_1 + \lambda S_2 = 0$ is a quadratic in $x$ and $y$ with equal coefficients of $x^2$ and $y^2$ (and vanishing $xy$ term), hence represents a circle (except for degenerate $\lambda$ values where circle reduces to a straight line).

Let $(x_0, y_0)$ be a point of intersection, so that $S_1(x_0, y_0) = 0$ and $S_2(x_0, y_0) = 0$. Then \[ S_1(x_0, y_0) + \lambda S_2(x_0, y_0) = 0 + \lambda \cdot 0 = 0 \] Therefore, every common point of $S_1=0$ and $S_2=0$ lies on every member of the family. For a particular value of $\lambda$, this gives a unique circle passing through both common points and one additional condition (such as passing through a third point).

Equation of the Radical Axis and Common Chord

Let two circles have equations $S_1 \equiv 0$ and $S_2 \equiv 0$. The equation of their common chord is obtained by subtracting their equations: \[ S_1 - S_2 = 0 \] Writing out explicitly, \[ (x^2 + y^2 + 2g_1x + 2f_1 y + c_1) - (x^2 + y^2 + 2g_2 x + 2f_2 y + c_2) = 0 \] \[ 2x(g_1-g_2) + 2y(f_1-f_2) + (c_1-c_2) = 0 \] This equation represents the radical axis or the chord common to both circles.

Family of Circles Touching a Fixed Line at a Fixed Point

Let $L \equiv 0$ be the equation of a fixed straight line, and $(x_1, y_1)$ a fixed point lying on the line $L=0$. The family of all circles tangent to $L=0$ at $(x_1, y_1)$ is given by: \[ (x - x_1)^2 + (y - y_1)^2 + \lambda L = 0 \] where $\lambda$ runs over all real numbers. This family is derived from the requirement that every member passes through $(x_1, y_1)$, and its tangent at this point coincides with $L=0$.

Family of Circles Touching a Fixed Circle and a Fixed Line at the Point of Contact

Let $S \equiv 0$ be the equation of a given circle, and $L \equiv 0$ the equation of a given tangent to $S=0$. The family of all circles touching $S=0$ and $L=0$ at their common point of contact is given by: \[ S + \lambda L = 0 \] where $\lambda$ is real. Each value of $\lambda$ provides a different circle tangent to $L$ at the same contact point as $S$.

Family of Circles Circumscribing a Triangle Whose Sides are Given Lines

Let the sides of a triangle be given by equations $L_1=0$, $L_2=0$ and $L_3=0$. The equation for the general circle passing through the intersection points of all three lines (i.e., the triangle's vertices) is: \[ L_1L_2 + \mu L_2L_3 + \nu L_3L_1 = 0 \] where $\mu, \nu$ are arbitrary real parameters. This equation includes all circles that pass through the triangle's three vertices, with particular parameter values giving each such circumcircle.

Criterion for Orthogonality of Circles in a Family

Two circles given by $S_1 \equiv x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $S_2 \equiv x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ intersect orthogonally if, at every point of their intersection, the tangents to both circles are perpendicular. The necessary and sufficient condition for orthogonality is: \[ 2(g_1g_2 + f_1f_2) = c_1 + c_2 \] This relation is derived using the gradients of the tangents at the point of intersection and the orthogonality condition for vectors.

Family of Concentric Circles

All circles sharing the same center $(h, k)$ are called concentric circles. Their general equations take the form: \[ (x - h)^2 + (y - k)^2 = r^2 \] or, equivalently, \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] where $g$ and $f$ are fixed ($g = -h$, $f = -k$), and $c$ is arbitrary. For further discussion on concentric circles and their intersections, refer to the Area Enclosed By Circle module.

Family of Circles with Centers Lying on a Given Line

If the locus of centers is constrained, such as all centers lying on $ax + by + c = 0$, the family is obtained by substituting the center $(h, k)$ into the line's equation: \[ a h + b k + c = 0 \] The circle equations are: \[ (x-h)^2 + (y-k)^2 = r^2 \] with parameters $h$, $k$ constrained by the above. Eliminating $k$ in terms of $h$ (or vice versa) and expressing $r$ as arbitrary parameter, the resulting equations represent the desired family.

Worked Example: Circle through Intersection of Two Given Circles and a Point

Given: Circles $S_1\!: x^2 + y^2 - 8x - 2y + 7 = 0$ and $S_2\!: x^2 + y^2 - 4x + 10y + 8 = 0$; required: Circle passing through their intersection and the point $(-1, -2)$.

The family of circles passing through the intersection is: \[ S_1 + \lambda S_2 = 0 \] Substitute the equations: \[ (x^2 + y^2 - 8x - 2y + 7) + \lambda(x^2 + y^2 - 4x + 10y + 8) = 0 \] \[ (x^2 + y^2) + \lambda(x^2 + y^2) - 8x - 2y + 7 - 4\lambda x + 10\lambda y + 8\lambda = 0 \] \[ (1+\lambda) x^2 + (1+\lambda) y^2 - (8 + 4\lambda)x + (-2 + 10\lambda) y + (7+8\lambda) = 0 \] The point $(-1, -2)$ lies on the required circle: \[ (1+\lambda)(-1)^2 + (1+\lambda)(-2)^2 - (8+4\lambda)(-1) + (-2+10\lambda)(-2) + (7 + 8\lambda) = 0 \] Calculate each term: \[ (-1)^2 = 1,\quad (-2)^2 = 4 \implies (1+\lambda)(1 + 4) = 5(1 + \lambda) \] \[ -(8 + 4\lambda)(-1) = 8 + 4\lambda \] \[ (-2 + 10\lambda)(-2) = 4 - 20\lambda \] \[ 7 + 8\lambda \] Sum: \[ 5(1+\lambda) + (8 + 4\lambda) + (4 - 20\lambda) + (7 + 8\lambda) = 0 \] \[ 5 + 5\lambda + 8 + 4\lambda + 4 - 20\lambda + 7 + 8\lambda = 0 \] Group like terms: \[ (5 + 8 + 4 + 7) + (5\lambda + 4\lambda -20\lambda + 8\lambda) = 0 \] \[ 24 + (-3\lambda) = 0 \] \[ -3\lambda = -24 \implies \lambda = 8 \] Substitute $\lambda = 8$ into main equation: \[ (1 + 8) x^2 + (1 + 8) y^2 - (8 + 4 \times 8) x + (-2 + 10 \times 8) y + (7 + 8 \times 8) = 0 \] \[ 9x^2 + 9y^2 - (8 + 32)x + (-2 + 80)y + (7 + 64) = 0 \] \[ 9x^2 + 9y^2 - 40x + 78y + 71 = 0 \]

Final result: The required equation is $9x^2 + 9y^2 - 40x + 78y + 71 = 0$.

Orthogonality and Intersection Angle in Family of Circles

For two general circles $S_1\!: x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ and $S_2\!: x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$, the cosine of the angle $\theta$ at which they intersect is expressed by: \[ \cos \theta = \frac{2(g_1g_2 + f_1f_2) - (c_1 + c_2)}{2 \sqrt{g_1^2 + f_1^2 - c_1}\, \sqrt{g_2^2 + f_2^2 - c_2}} \] For orthogonality $(\theta = 90^\circ)$, the numerator must be zero.

Related Concepts

For additional theory, see the Common Tangents and Chord Length Formula references. For 3D extensions, differences, and areal properties, refer to Difference Between Circle And Sphere and Area Of A Circle Formula.