Circle Through the Intersection of Two Circles

We will learn how to find the equation of a circle through the intersection of two given circles.

The equation of a family of circles passing through the intersection of the circles P1 = x2 + y2 + 2g1x + 2f1y + c1 = 0 and P2 = x2 + y2 + 2g2x + 2f2y + c2 = 0 is P1 + λP2 = 0 i.e., (x2 + y2 + 2gx1 + 2fy1 + c1) + λ(x2 + y2 + 2g2x + 2f2y + c2) = 0, where λ (≠ -1) in an arbitrary real number.

Proof:

Let the equations of the given circles be 

P1 = x2 + y2 + 2g1x + 2f1y + c1 = 0 ………………………..(i) and

P2 = x2 + y2 + 2g2x + 2f2y + c2 ………………………..(ii)

Circle Through the Intersection of Two CirclesCircle Through the Intersection of Two Circles

Consider the equation P1 + λP2 = 0 i.e., the equation of any curve through the points of intersection of the circles (1) and (2) is

(x2 + y2 + 2g1x + 2f1y + c1) + λ(x2 + y2 + 2g2x + 2f2y + c2) = 0 ………………………..(iii)

Clearly, it represents a circle for all values of λ except λ = -1. For λ = -1 (iii) becomes a first degree equation in x, y which represents a line. In order to prove that it passes through the points of intersection of the two given circles, it is sufficient to show that their points of intersection satisfy (iii).

Let (x1, y1) be a point of intersection of the given circles.

Then,
x21+y21+2g1x1+2f1y1+c1 and x21+y21+2g2x1+2f2y1+c2

⇒ (x21+y21+2g1x1+2f1y1+c1) +  λ(x21+y21+2g2x1+2f2y1+c2) = 0 + λ0 = 0

⇒ (x1, y1) lies on (iii).

Similarly, it can be proved that the second point of intersection of the given circles also satisfy (i)

Hence, (iii) gives the family of circles passing through the intersection of the given circles.

In other words, the equation of any curve through the points of intersection of the circles (i) and (ii) is
(x2 + y2 + 2g1x + 2f1y + c1) + λ(x2 + y2 + 2g2x + 2f2y + c2)………………………..(iv)

⇒ (1 + λ)(x2 + y2) + 2(g1 + g2λ)x + 2(f1 + f2λ)y + c1 + λc2 = 0

⇒ x2 + y2 + 2 ∙ g1+g2λ1+λ x + 2 ∙ f1+f2λ1+λy + c1+c2λ1+λ = 0 ………………………..(v)   

If λ ≠ - 1, then equation (v) will represent the equation of a circle. Therefore, the equation (iv) represents the family of circles through the points of intersection of the circles (1) and (2). 


Solved examples to find the equations of a circle through the points of intersection of two given circles: 

1. 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 passes through the point (-1, -2).

Solution:

The equation of any circles passing through the intersection of the circles S1 = x2 + y2 - 8x - 2y + 7 = 0 and S2 = x2 + y2 - 4x + 10y + 8 = 0 is S1 + λS2 = 0 

Therefore, the equation of the required circle is (x2 + y2 - 8x - 2y + 7) + λ(x2 + y2 - 4x + 10y + 8) = 0, where λ (≠ -1) in an arbitrary real number

This circle passes through the point (-1, -2), 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 (x2 + y2 - 8x - 2y + 7) + λ(x2 + y2 - 4x + 10y + 8) = 0 we get the required equation as 9x2 + 9y2 – 40x + 78y + 71 = 0.


2. Find the equation of the circle through the intersection of the circles x2 + y2 - x + 7y - 3 = 0 and x2 + y2 - 5x - y + 1 = 0, having its centre on the line x + y = 0. 

Solution:

x2 + y2 - x + 7y - 3 + λ(x2 + y2 - 5x - y + 1) = 0, (λ ≠1)

⇒(1 + λ) (x2 + y2) - (1 +  5λ)x + (7 - λ)y - 3 + λ = 0

⇒ x2 + y2 - 1+5λ1+λx - λ71+λy + λ31+λ = 0 …………….(i)

Clearly, the co-ordinates of the centre of the circle (i) are [1+5λ2(1+λ), λ72(1+λ)] By question, this point lies on the line x + y = 0. 

Therefore, 1+5λ2(1+λ) + λ72(1+λ) = 0 

⇒1 + 5λ + λ - 7 = 0 

⇒ 6λ =  6

⇒ λ = 1

Therefore, the equation of the required circle is 2(x2 + y2) - 6x + 6y - 2 = 0, [putting λ = 1 in (1)] 

⇒ x2 + y2 - 3x + 3y - 1 = 0. 

 The Circle




11 and 12 Grade Math 

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