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Question

Find the equation of family of circles passing through the point of intersection of the circles x2 + y2 2x 4y 4 = 0 and x2 + y2 10x 12y + 40 = 0 and whose radius is 4.


A

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C

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Solution

The correct options are
B


C


Family of circles through intersection of circles is given by

S1+λS2=0

(x2+y22x4y4)+λ(x2+y210x12y+40)=0

(1+λ)x2+(1+λ)y22(1+5λ)x2(2+6λ)y+40λ4=0

x2+y22(1+5λ)1+λx+2(2+6λ)1+λy+40λ41+λ=0 - - - - - - (1)

center of the circle (-g,-f)

(1+5λ1+λ,2+6λ1+λ)

Given radius of the required circles is =4

g2+f2c=4

(1+5λ1+λ)2+(2+6λ1+λ)240λ41+λ=16

16(1+λ2+2λ)=1+10λ+25λ2+4+24λ+36λ240λ2+4+4λ

16+332λ+16λ2=21λ22λ+9

5λ234λ7=0

(λ7)(5λ+1)=0

λ=7, 15

Substituting these values of λ in equation (1)

The required circles are

when λ=7

x2+y22(1+5×7)1+72(2+6×7)1+7+40×741+7=0

X2+y29x11y+692=0

When λ=15

x2+y22(1+5(15))1152(2+6(15))115+40(15)4115

x2+y202y15=0

x2+y22y15=0

Requiredequations of family of circle are 2x2+2y218x22y+69=0 and x2+y22y15=0


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