Family of circles is one of the important topics in coordinate geometry. A large collection of circles is called a family of circles. Circles having the same properties are combined together in various ways to form the family of circles. In this article, we will learn the formulas for the family of circles, along with solved examples.

The general equation of a circle is given by x² + y² + 2gx + 2fy + c = 0. We will discuss some of the ways of finding the family of circles under the given certain conditions.

Let S = x² + y² + 2gx + 2fy + c = 0

S₁ = x² + y² + 2g₁x + 2f₁y + c₁ = 0

S₂ = x² + y² + 2g₂x + 2f₂y + c₂ = 0

L = lx + my + n = 0

1. Family of circles having a fixed centre

The equation is given by (x-h)² + (y-k)² = r².

Here (h, k) is fixed and r is a varying parameter. This is the equation of the family of concentric circles. Fixation of the radius will give a particular circle.

2. Family of circles passing through the point of intersection of line and circle

The required equation for the family of circles passing through the point of intersection of circle S = 0 and line L = 0 is given by S+λL = 0, where λ is a parameter.

3. Family of circles passing through the point of intersection of two circles

The equation of the family of circles passing through the point of intersection of two circles S₁ = 0 and S₂ = 0 is given by S₁ + λS₂ = 0. Where λ ≠ -1.

Note that both circles should be in the given format of S₁ and S₂.

If λ = -1, the above equation becomes

S₁ – S₂ = 0, which is the equation of the common chord if the circles are intersecting.

4. Family of circles passing through two given points

The equation of the family of circles passing through two given points P(x₁, y₁) and Q(x₂, y₂) is given by

5. Family of circles touching a line at a given point

The equation of family of circles touching the line y-y₁ = m(x-x₁) at a point (x₁, y₁) on any finite m is given by

(x-x₁)² + (y-y₁)² + λ[(y-y₁) – m(x-x₁)] = 0

If m is infinite, the equation becomes (x – x₁)² + (y-y₁)² + λ(x-x₁) = 0

Types of Circles

Orthogonal Circles

If two circles cut each other at right angles, they are called orthogonal circles. Two circles of radii r₁ and r₂ and if d is the distance between their centres, then they are orthogonal if

r₁²+r₂² = d².

Let S₁ = x² + y² + 2g₁x + 2f₁y + c₁ = 0

S₂ = x² + y² + 2g₂x + 2f₂y + c₂ = 0

S₁ and S₂ are orthogonal if 2g₁g₂ + 2f₁f₂ = c₁+ c₂

Concentric Circles

Circles having the same centre and different radius are called concentric circles. Consider a circle with centre (-g, -f) and radius √(g²+f²-c). The equation of the circle is given by

x²+y²+2gx+2fy+c = 0

Equation of a concentric circle to above circle is given by x²+y²+2gx+2fy+c₁ = 0

Both circles will have the same centre (-g, -f) but a different radius. (c ≠ c₁)

Contact of Circles

Case 1: If the sum of the radii of two circles is equal to the distance between the centres, then the circles touch externally. The circles will satisfy the condition

c₁c₂ = r₁ + r₂

Case 2: If the difference between the radii and the distance between the centres are equal, then the circles touch internally. The circles will satisfy the condition

c₁c₂ = r₁ – r₂

Also Read

JEE Circles and System of Circles Previous Year Questions with Solutions

JEE Main Circle Previous Year Questions with Solutions

Solved Examples

Question 1: Let the circles be S₁ = x² + y² – 8x + 6y – 23 = 0 and S₂ = x² + y² – 2x – 5y + c = 0. Find the value of c if the circles S₁ and S₂ are orthogonal.

(a) 16

(b) -16

(c) 30

(d) 23

Solution:

Given S₁ = x² + y² – 8x + 6y – 23 = 0

S₂ = x² + y² – 2x – 5y + c = 0

Comparing with general equation we get

g₁ = -4, f₁ = 3, c₁ = -23

g₂ = -1, f₂ = -5/2, c₂ = c

S₁ and S₂ are orthogonal.

So 2g₁g₂ + 2f₁f₂ = c₁ + c₂

=> 2×-4×-1 + 2×3×-5/2 = -23 + c

=> 8 – 15 = -23 + c

=> c = -7 + 23

=> c = 16

Hence, option a is the answer.

Question 2: Find the equation of a circle w passing through Q(1, 2) and touching 2x+y = 5 at point P(2, 1).

(a) x² + y² + 5 = 0

(b) x² + y² – 5 = 0

(c) x² + y² – 8x – 5 = 0

(d) x² – y² – 8x – 5 = 0

Solution:

The equation of point circle at P(2,1) is (x-2)² + (y-1)² = 0

Given line is 2x+y = 5

Equation of w => (x-2)² + (y-1)² + λ(2x+y-5) = 0

Since w passes through (1, 2)

=> (1-2)² + (2-1)² + λ(2+2-5) = 0

=> 2 – λ = 0

=> λ = 2

So the required equation is (x-2)² + (y-1)² + 2(2x+y-5) = 0

=> x² – 4x + 4 + y² – 2y + 1 + 4x + 2y – 10 = 0

=> x² + y² – 5 = 0

Hence, option b is the answer.

Question 3: The differential equation of family of circles having the centre at origin is

(a) x + y (dy/dx) = 0

(b) x – y (dy/dx) = 0

(c) -x + y (dy/dx) = 0

(d) none of these

Solution:

General equation of circle is (x-h)² + (y-k)² = r²

(h, k) is the centre and r is the radius of the circle.

Given centre is (0, 0)

So the equation becomes x² + y² = r²

Differentiate w.r.t.x

=> 2x + 2y (dy/dx) = 0

=> x + y (dy/dx) = 0 is the required equation.

Hence, option a is the answer.

Question 4: Find the equation of the circle passing through the point of intersection of the circle x² + y² = 4 and the line 2x + y = 1 and having minimum possible radius.

(a) 5x² + 5y² – 4x – 2y – 18 = 0

(b) 5x² + 5y² + 6x – 18y – 15 = 0

(c) 5x² + 5y² + 4x + 9y – 5 = 0

(d) 5x² + 5y² + 18x +6y – 5 = 0

Solution:

The equation for the family of circles passing through the point of intersection of circle S = 0 and line L = 0 is given by S+λL = 0

Given S = x² + y² = 4

L = 2x+y = 1

So the equation is x² + y² – 4 + λ(2x+y-1) = 0…(i)

Comparing with general equation, we get

g = λ, f = λ/2, c = -4 – λ

Radius = √(g²+f² -c)

= √( λ² + (λ²/4) +4+ λ)

= √((5λ²/4)+4+ λ)

= √((5λ²+16+4λ)/4)

Radius has to be minimum.

Let f(λ) = (5λ²+16+4λ)/4)

f’(λ) = (10λ+4)/4

f’’(λ) = 10/4 = 5/2 > 0

For minimum, f’(λ) should be zero and f’’(λ)>0

=> (10λ+4)/4 = 0

=> 10λ = -4

=> λ = -4/10 = -⅖

Put λ in (i)

x² + y² – 4 – (⅖)(2x+y-1) = 0

=> 5x² + 5y² – 4x – 2y – 18 = 0

Hence, option a is the answer.

Question 5: 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 passing through the point (-1, -2).

(a) 9x² – 9y² – 40x + 78y + 72 = 0

(b) 9x² + 9y² – 40x + 78y + 71 = 0

(c) 4x² + 4y² – 40x + 78y + 71 = 0

(d) 9x² + y² – 40x – 78y + 71 = 0

Solution:

The equation of family of circles passing through the point of intersection of two circles S₁ = 0 and S₂ = 0 is given by S₁ + λS₂ = 0

Given S₁ = x² + y² – 8x – 2y + 7 = 0

S₂ = x² + y² – 4x + 10y + 8 = 0

=> x² + y² – 8x – 2y + 7 + λ(x² + y² – 4x + 10y + 8) = 0

Since it passes through (-1, -2)

=> 1 + 4 + 8 + 4 + 7 + λ(1 + 4 + 4 – 20 + 8)=0

=> 24 – 3λ = 0

=> λ = 24/3 = 8

Required equation is x² + y² – 8x – 2y + 7 + 8(x² + y² – 4x + 10y + 8) = 0

=> 9x² + 9y² – 40x + 78y + 71 = 0

Hence, option b is the answer.

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