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Question

Find the equation of family of circles through the intersection of x2 + y2 − 6x + 2y + 4 = 0 and x2 + y2 + 2x − 4y − 6 = 0 whose center lies on y = x.


A

x2+y2107x107y127=0

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B

x2+y2+107x+107y127=0

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C

x2+y2107x+107y+127=0

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D

x2+y2+107x+107y+127=0

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Solution

The correct option is A

x2+y2107x107y127=0


Given circles

S1=0 x2+y26x+2y+4=0 - - - - - - -(1)

S2=0 x2+y22x4y6=0 - - - - - - -(2)

The equation of family of circles passing through the points of intersection of two circles S1 =0 & S2=0 is S1+λS2=0

(x2+y26x+2y+4)+λ(x2+y2+2x4y6)=0 - - - - - - - (3)

(1+λ)x2+(1+λ)y2+(2λ6)x+(24λ)y+46λ=0

Dividing both sides by 1+λ

x2+y2+2(λ3)1+λx+2(12λ)1+λy+46λ1+λ=0

Coordinates of center of circle is (-g,-f)

(3λ1+λ,2λ11+λ)

Since, center lies on the line y=x

3λ1+λ=2λ11+λ

{λ1}

λ=43

Equation of family of circles passing through the point of intersection of circle is

(x2+y26x+2y+4)+43(x2+y2+2x4y6)=0

73x2+73y2103x103y4=0

x2+y2107x107y127=0

Option A is correct


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