Family of Circles Passing Through Two Points

In geometry, a family of circles passing through two fixed points consists of all the circles that can be drawn through these points. This concept is particularly useful in coordinate geometry and can be applied to solve various problems involving circles and points.

Understanding the Concept

Given two points ( P(x_1, y_1) ) and ( Q(x_2, y_2) ), there are infinitely many circles that pass through both points. These circles will have different radii and positions, but they will all share the common property of intersecting at ( P ) and ( Q ).

General Equation of a Circle

The general equation of a circle in the Cartesian plane with center ( (h, k) ) and radius ( r ) is given by:

[ (x - h)^2 + (y - k)^2 = r^2 ]

Family of Circles Through Two Points

To find the equation of the family of circles passing through two points, we use the concept that any circle passing through ( P ) and ( Q ) can be expressed as a linear combination of the equations of two known circles that also pass through ( P ) and ( Q ).

Let's consider two circles with the following equations:

1. Circle ( C_1 ) with equation ( (x - x_1)^2 + (y - y_1)^2 = r_1^2 ) passing through ( Q ).
2. Circle ( C_2 ) with equation ( (x - x_2)^2 + (y - y_2)^2 = r_2^2 ) passing through ( P ).

The equation of the family of circles passing through ( P ) and ( Q ) can be written as:

[ (x - x_1)^2 + (y - y_1)^2 + \lambda[(x - x_2)^2 + (y - y_2)^2] = 0 ]

where ( \lambda ) is a parameter that varies for different members of the family.

Important Points and Differences

Aspect	Description
Fixed Points	The two points ( P ) and ( Q ) through which all circles in the family must pass.
Parameter ( \lambda )	A real number that changes the size and position of the circle within the family.
Center	The center of each circle in the family will lie on the perpendicular bisector of the line segment ( PQ ).
Radius	The radius of each circle will vary depending on the value of ( \lambda ).

Examples

Example 1: Finding the Family of Circles

Given two points ( P(2, 3) ) and ( Q(4, 5) ), find the equation of the family of circles passing through both points.

Solution:

We can use the general form of the family of circles:

[ (x - 2)^2 + (y - 3)^2 + \lambda[(x - 4)^2 + (y - 5)^2] = 0 ]

This equation represents the family of circles passing through ( P(2, 3) ) and ( Q(4, 5) ) for different values of ( \lambda ).

Example 2: Specific Circle in the Family

Find the equation of a specific circle in the family that passes through ( P(1, 2) ) and ( Q(3, 4) ) and has its center on the x-axis.

Solution:

The center of the circle must have coordinates ( (h, 0) ). Since the center lies on the perpendicular bisector of ( PQ ), we can find ( h ) by averaging the x-coordinates of ( P ) and ( Q ):

[ h = \frac{x_1 + x_2}{2} = \frac{1 + 3}{2} = 2 ]

Now, we can write the equation of the family of circles:

[ (x - 1)^2 + (y - 2)^2 + \lambda[(x - 3)^2 + (y - 4)^2] = 0 ]

To find the specific value of ( \lambda ) for our circle, we substitute ( (h, 0) ) into the equation and solve for ( \lambda ):

[ (2 - 1)^2 + (0 - 2)^2 + \lambda[(2 - 3)^2 + (0 - 4)^2] = 0 ] [ 1 + 4 + \lambda[1 + 16] = 0 ] [ 5 + 17\lambda = 0 ] [ \lambda = -\frac{5}{17} ]

The equation of the specific circle is:

[ (x - 1)^2 + (y - 2)^2 - \frac{5}{17}[(x - 3)^2 + (y - 4)^2] = 0 ]

This example illustrates how we can find a specific member of the family of circles passing through two points by incorporating additional conditions.