Families of Circles

A collection of circles is called a family or a system of circles.

Let S and S' be two intersecting (or touching) circles, then S +k S' = 0, k -1, represents a family of circles through their points (or point) of intersection.

Remarks

1. If k = -1, then the equation S +kS' = 0 reduces to S -S' = 0 which represents the common chord in case of intersecting circles or common tangent in case of touching circles.
2. The equation S +kS = 0 represents all members of the family except the member S'. If we need the member S', then take the equation of the family as S' +kS = 0.

Illustrative Examples

Example

Find the equation of the family of circles passing through the points A(0,4) and B(0,-4).

Solution

Let the equation of the desired family of circles be
x² +y² +2gx +2fy +c = 0     ...(i)
As all these circles pass through the points A(0,4) and B(0,-4), we get
0 +16 +0 +8f +c = 0 => 8f +c +16 = 0               ...(ii),
0 +16 +0 -8f +c = 0 => -8f +c +16 = 0   ...(iii)
On solving (ii) and (iii), we get f = 0, c = -16.
Substituting these values in (i), we get
x² +y² +2gx -16 =0 ...(iv)
Note that for every real value of g,
g² +f² -c = g² +16 > 0, therefore, (iv) represents a circle.
Hence the equation x² +y² +2gx -16 = 0 for different real values of g represents the desired family of circles. It is one-parameter family of circles where g is the parameter.

Example

Find the equation of the circle which passes through the points of inter-section of x² +y² -4 = 0 and x² +y² -2x -4y +4 = 0 and touches the line x +2 y = 0.

Solution

The equations of the given intersecting circles are
S = x² +y² -4 = 0, and
S' = x² +y² -2 x -4y +4 = 0
The equation of the common chord of these circles is
S -S' = 0   => l = 2 x +4 +4 y -8 = 0
The equation of the family of circles passing through the intersection of the given circles is
x² +y² -4 +k (2x +4 +4y -8) = 0 ...(i)
Its center is (-k,-2k) and
radius = [k² +4k² +4 +8k] = [5k² +8k +4].
For the particular member of the family which touch the line x +2y = 0, we have
|-k +2 (-2k)|/ [1² +2²] = [5k² +8k +4]
=>   5|k|/5 = [5k² +8k +4]
=> 5 k² = 5 k²+8 k +4
=> 8 k +4 = 0 => k = -1/2
Substituting this value of k in (i), the equation of the required circle is
x² +y² -4 - (1/2)(2x +4y -8) = 0 i.e. x² + y² -x -2y = 0.

Exercise

1. Find the equation of the family of circles passing through the origin.
2. Find the equation of the family of circles passing through the origin and the point (0,1).
3. Find the equation of the family of circles passing through the points A(5,0) and B(-5,0).
4. Find the equation of the family of circles with radius 3 and whose centers lie on the x-axis.
5. Find the equation of the family of concentric circles with center as (-4,2). Also find a member of the family which touches the line x -y = 3.
6. Show that the equation of the family of circles which touch both the co-ordinate axes can be put into the form x² +y² ±2rx ±2ry +r² = 0, where r is radius.
   [Hint. Let (, ) be center of the family. Since it touches both the axes
   i.e. y = 0 and x = 0, so || = ||   r  => = ±r, = ±r.]
7. Find the equation of the circle which passes through the origin and the points of intersection of the circles x² +y² +2x +2y -2 = 0 and x² +y² +4x -8y +4 = 0.
8. Find the equation of the circle through the points of intersection of the circles x² +y² +2x +3y -7 = 0 and x² +y² +3x -2y -1 = 0 and through the point (1,2).
9. Find the equation of the circle which passes through the point (1,-1) and through the points of intersection of the circles x² +y² +2x -2y -23 = 0 and 3x² +3y² +12x -4y -9 = 0.
10. Find the equation of the circle passing through the point (2,3) and through the points of intersection of the circle x² +y² +3x -4y +5 = 0 and the line x -y +2 = 0.
11. Find the equation of the circle through the intersection of the circles x² +y² -8x -2y +7 = 0 and x² +y² -4x +10y +8 = 0 and having its center on the x-axis.

Answers