Family of Circles for IIT JEE

A circle is a closed shape in which all the points lying on the surface of the circle joining its center are equal and known as the radius of the circle. A huge collection of circles is called a Family of Circles. There are various types of circles available around us. The Family of Circles is one of the most important parts of coordinate geometry in which different circles are present at the same time, through which the calculations to find the values in circles becomes easy. The circles consisting of the same properties are combined in various ways to form the Family of Circles. The general equation of the circle can be given as x2 + y2 + 2gx + 2fy + c = 0. However, the equation contains three unknown figures i.e. g, f and c so we require at least three conditions to get a unique circle.

For instance, if we are given two circles and we need to resolve the third circle touching the rest of both the circles. For this, we need one more condition. Without the condition, we get the equation of the Family of Circles that satisfies the given two conditions. The 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 certain conditions:

1. When Family of Circles Have a Fixed Center

The equation is (x-y)2 + (y-k)2 = r2

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

2. Equation of Family of Circles passing through the intersection of two circles S1 = 0 and S2 = 0.

The general equation of the Family of Circles is passing through the intersection of S1 and S2 which is given by S1 + KS2 = 0, where k ≠ -1. Again we are left with one parameter (k) equation of the Family of Circles. Further, the parameter value of the parameter gives a unique circle.

Caution:

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

Let \[S_{1} \equiv x_{2} + y_{2} + 2g_{1}x + 2f_{1}y + c_{1} = 0 \]

\[S_{2} \equiv x_{2} + y_{2} + 2g_{2}x + 2f_{2}y + c_{2} = 0 \]

Since, the point lies on both the circles,

\[\Rightarrow xA_{2} + yA_{2}+ 2g_{1}xA + 2f_{1}yA + c_{1} = 0\]

So, \[\Rightarrow xA_{2} + yA_{2}+ 2g_{2}xA + 2f_{2}yA + c_{2} = 0\]

\[\Rightarrow xA_{2} + yA_{2}+ 2g_{1}xA + 2f_{1}yA + c_{1} + \lambda (xA_{2} + yA_{2}+ 2g_{1}xA + 2f_{1}yA + c_{1}) = 0\]

\[\Rightarrow\] Point A (xA, yA) lies on \[S_{1} + \lambda S_{2} = 0 ∀ λ ∈ R \]

Similarly point B (xB, yB) lies on S1 + λ S2 = 0 ∀ λ ∈ R

Therefore, S1 + λ S2 = 0 is the Family of Circles through the intersection of S1 = 0 and S2 = 0.

3. Equation of the circle circumscribing a triangle whose sides are given by L1 = 0, L2 = 0 and L3 = 0.

This equation is given by L1L2 + λL2L3 + µL3 L1 = 0, which provides a coefficient of x y= 0 and coefficient of x2 = coefficient of y2. Moreover, the particular value of the parameters λ and µ gives you a unique circle.

4. Family of Circle Touching the Circle S= 0 and the Line L = 0 at their Point of Contact

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

5. Family of a circle passing through two mention points i.e. A(x1, y1) and B(x2, y2).

The equation of the circle is (x – x1) (x – x2) + (y – y1) (y – y2) = 0 having AB is its diameter and \[\begin{bmatrix}x & y & 1 \\ x_{1} & y_{1}c & 1 \\ x_{2} & y_{2} & 1 \end{bmatrix} = 0\] is the equation of a line passing through points A and B. 

The required equation of the Family of Circles is \[(x−x_{1})(x−x_{2}) + (y−y_{1})(y−y_{2}) + \lambda \begin{bmatrix}x & y & 1 \\ x_{1} & y_{1}c & 1 \\ x_{2} & y_{2} & 1 \end{bmatrix} = 0\] 

6. Family of Circles Touching the Given Line L = 0 at a Point (x₁, y₁) on the Line is

(x – x1)2 + (y – y1)2 + λL = 0 is the particular value of the parameters λ which gives a unique circle. The equation of the Family of Circles touches the line y – y1 = m (x – x1) at (x1, y1) for any values of m is (x – x1)2 + (y – y1)2 + λ[(y – y1) – m(x – x1)] = 0. 

Here, we can have two sub-cases according to whether the line is parallel to The angle of The required center or y-axis.

If this line through (x1, y1) is parallel to x-axis, the equation of the Family of Circles touching it at (x1, y1) becomes (x – x1)2 + (y – y1)2 + K(y – y1) = 0.

Moreover, if the line through (x1, y1) is parallel to x-axis, then the equation of the Family of Circles touching it at (x1, y1) becomes (x – x1)2 + (y – y1)2 + K(y – y1) = 0.

7. Equation of Circle Circumscribing a Quadrilateral Whose Sides in order are Expressed by the Lines L1 = 0, L2 = 0, L3 = 0 and L4 = 0.

The equation of the required family is given by L1L3 + λL2L4 = 0 which provides a coefficient of x2 = coefficient of y2 and coefficient of XY = 0.

Common Chord of Two Circles:

The common chord is the one that joins the same point of intersection of two circles. Consider the circle equations:

\[S_{1} \equiv x_{2} + y_{2} + 2g_{1}x + 2f_{1}y + c_{1} = 0 \]

\[S_{2} \equiv x_{2} + y_{2} + 2g_{2}x + 2f_{2}y + c_{2} = 0 \]

Then the common chord equation will be \[S_{1} - S_{2} = 0\].

Therefore, this is the equation: \[2x (g_{1} - g_{2}) + 2y (f_{1} - f_{2}) + c_{1} – c_{2} = 0\].

Point to Remember:

• Consider the length of the common chord to be : \[2 \sqrt{(r_{1}^{2} – p_{1}^{2})} = 2\sqrt{(r_{2}^{2} – p_{2}^{2})}\].

• Here, p₁ and p₂ are lengths of the perpendicular bisector drawn from the center of the chord.

• The coefficients of x₂ and y₂ should be the same in both equations.

• If the length of the common chord is zero when two circles touch each other,

• The diameter of the smaller circle is equal to the maximum length of the common.

The Angle of The Intersection of Two Circles:

The angle of intersection is defined as an angle between their tangents at their point of intersection. Let it be S=0 and S’=0.

Consider \[S“ \equiv x_{2} + y_{2} + 2g_{2}x + 2f_{2}y + c_{2} = 0 \], and \[S“ \equiv x_{2} + y_{2} + 2g_{1}x + 2f_{1}y + c_{1} = 0 \] and their radii are r₁, r₂.

The ‘d’ distance between their centers.

Thus the θ i.e. the angle of intersection:

\[cos θ = \frac{(r_{1}^{2} + r_{2}^{2} – d^{2})}{2r_{1}r_{2}}\]  or   \[cos θ = \frac{2(g_{1}g_{2} + f_{1}f_{2}) – (c_{1} + c_{2})}{ 2\sqrt{(g_{1}^{2} + f_{1}^{2} – c_{1})} \sqrt{(g_{2}^{2} + f_{2}^{2} – c_{2})}}\].

Remember:

• If the perspective of the intersection of the circles, i.e. the perspective among their tangents on the factor of intersection, is \[90^{\circ}\], the 2 circles are stated to intersect orthogonally.

• 2gg₁ + 2ff₁ = c + c₁ is the situation for the 2 circles to reduce every different orthogonality, where (–g, –f) and (–g_{1}, –f_{1}) are the respective circle centers, S = zero and S₁ = zero.

• When circles meet orthogonally, the length of 1 circle's tangent from the opposite circle's center equals the radius of the opposite circle.

Types of Circles

1. Concentric Circles

When two or more circles have a common center, then these circles are called concentric circles.

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In the above figure, there are three circles inside one another. All these circles are of different sizes and have different radii. All circles have one common center, i.e. C but not the common radius. Like in this given figure, the first inner circle has radius1, the middle circle has radius2 and the third outer circle has radius3. It means that all the three circles have different measurements having the same center. Consequently, if all the circles have a radius then the circles will not be concentric. As a result, they will lie to each other and will not be able to see and treat each other as one single circle.

Here is the equation of the concentric circle – x2 + y2 + 2gx + 2fy + c = 0 is x2 + y2 + 2gx + 2fy + k = 0 (Equation differs only by the constant term).

2. Contact of Circles

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

There may be two cases in contact with circles. 

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

Equation – c1c2 = r1 + r2

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In the above figure, there are two circles touching each other externally at point P. Both circles have different centers c1 and c2. The distance between both centers is the sum of their radii.

Case (ii) when two circles touching the outer surface internally having their centers are equal to the difference of their radii. 

Equation – c1c2 = r1 – r2 or r2 - r1

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In the above figure, the outer of the two circles are touching internally at the P point. Similarly, in case (i), both circles have different centers c1 and c2. 

3. Orthogonal Circles

When two circles cut each other at right angles, they are called orthogonal circles. According to the Pythagoras theorem, two circles of radii r1 and r2 whose center area is distanced apart are orthogonal, if r12 + r22 = d2. Figure as an example is given below. 

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

Example: Find the equation of the circle through the intersection of the circles x22 + y22 - 8x - 2y + 7 = 0 and x22 + y22 - 4x + 10y + 8 = 0 and pass through the point (-1, -2).

Solution - The equation of any circles passing through the intersection of the circles S11= x22 + y22 - 8x - 2y + 7 = 0 and S22 = x22 + y22 - 4x + 10y + 8 = 0 is S11 + λS22 = 0 

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 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.