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Question

Find the equation of circle which touches 2x y + 3 = 0 and pass through the points of intersection of the line x + 2y 1 = 0 and the circle x2 + y2 2x + 1 = 0


A

x2 + y2 + 4y 1 = 0

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B

x2 + y2 4x 4y + 3 = 0

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C

x2 + y2 + 4x 1 = 0

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D

x2 + y2 + 4x + 4y + 3 = 0

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Solution

The correct option is B

x2 + y2 4x 4y + 3 = 0


Equation family of circle is

S+λL=0

x2+y22x+1+λ(x+2y1)=0

x2+y2x(xλ)+2λy+(1λ)=0 - - - - - -(1)

centre of circle is (-g,-f)

=(2λ2,λ)

And radius of the circle is =g2+f2c=(2λ2)2+λ2(1λ)

=124+λ24λ+4λ24+4λ

=125λ2=52|λ|

Since, circle touches the line 2xy+3=0, therefore perpendicular from the centre should be equal to

radius.

Centre (2λ2,λ)

Perpendicular distance from a point to a line is

∣ ∣2.((2λ)2)(λ)+34+1∣ ∣=52|λ|

|2.λ+λ+3|5

55=52|λ|

5=52|λ|

|λ|=2

λ=±2

Substituting the value of λ in equation (1)

When λ=2

x2+y2x(22)+2(2)y+(12)=0

x2+y2+4y1=0

When λ=2

x2+y2x(2+2)+2(2)y+(1+2)=0

x2+y24x4y+3=0

so,option A and B are correct.


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