For instance, if we are given two circles and we need to resolve the third circle touching the rest both the circles. For this, we need one more condition. Without the condition we get the equation of family of circles which satisfies the given two conditions. Imposition of a third condition will result in the equation which represents a particular circle.

Let’s discuss some of the ways of finding the family of the circle when you are given with certain conditions:

The equation is (x-y)

Where (h, k) is fixed and the only parameter that is varying is r. The fixation of the radius will give a particular circle.

The general equation of family of circles is passing through the intersection of S

Caution:

If k = –1, we will get equation of common chord i.e. straight line instead of circle. Let S

S

Since, point lies on both the circles,

⇒ x

So, x

⇒ x

⇒ Point A (x

Similarly point B (x

Therefore, S1 + λ S

This equation is given by L1L2

The family is given by equation S + λL = 0, where λ is a required family.

The equation of the circle is (x – x

is the equation of a line passing through the points A and B.

Required equation of family of circles is

x – x

Here, we can have two sub-cases according as whether the line is parallel to x -axis or y-axis.

If this line through (x

Moreover, if the line through (x

The equation of the required family is given by

When two or more circles have the common centre, then these circles are called concentric circles.

In the above figure, there are three circles inside one other. All these circles are of different size and are having different radius. All circles have one common centre i.e. C but not the common radius. Like in this given figure, the first inner circle is having radius_{1}, middle circle is having radius_{2} and third outer circle is having radius_{3}. It means that all the three circles have different measurement having the same centre. Consequently, if all the circles have the radius then the circles will not be concentric. As a result they will lie on each other and will not be able to see and treat each other as one single circle.

Here is the equation of concentric circle – x^{2 }+ y^{2 }+ 2gx + 2fy + c = 0 is x^{2 }+ y^{2 }+ 2gx + 2fy + k = 0 (Equation differ only by the constant term).

Here is the equation of concentric circle – x

When outer surface of two circles are touching, it is known as contact of circles.

There may be two cases in contact of circles.

Case (i) when two circles touching the outer surface externally and where the distance between their centres is equal to the sum of their radii. In this case, circles must satisfy the given equation.

Equation – c

In the above figure, there are two circles touching each other externally at point P. Both circles have different centres c_{1} and c_{2}. The distance between both centres is the sum of their radii.

Case (i) when two circles touching the outer surface internally having their centres is equal to the difference of their radii.

Equation – c

In the above figure, outer of the two circles are touching internally at P point. Similarly in case (i), both circles have different centres c_{1} and c_{2}.

When two circles cut each other at right angles, they are called orthogonal circles. According to the Pythagoras theorem, two circles of radii r

We are mentioning solved example below to find the equation of a circle through the points of intersection of two given circles.

Therefore, the equation of the required circle is (x22 + y22 - 8x - 2y + 7) + λ(x22 + y22 - 4x + 10y + 8) = 0, where λ (≠ -1) in an arbitrary real number

The point (-1, -2) has been passed through the circle. Therefore, (1 + λ) + 4(1 + λ) + 4(2 + λ) + 4(1 - 5λ) + 7 + 8λ = 0

⇒ 24 - 3λ = 0

⇒ λ = 8

Now putting the value of λ = 8 in the equation (x22 + y22 - 8x - 2y + 7) + λ(x22 + y22 - 4x + 10y + 8) = 0, we get the required equation as 9x22 + 9y22 – 40x + 78y + 71 = 0.