    Question

# The equation of the circle through the points of intersection of x2+y2−1=0,x2+y2−2x−4y+1=0 and touching the line x + 2y = 0, is

A

x2+y24x4y3=0

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B

3(x2+y2)+4x4y3=0

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C

x2+y2x2y=0

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D

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

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Solution

## The correct option is C x2+y2−x−2y=0 Family of circles is x2+y2−2x−4y+1+λ(x2+y2−1)=0. (1+λ)x2+(1+λ)y2−2x−4y+(1−λ)=0 x2+y2−21+λx−41+λy+1−λ1+λ=0 ............(i) Centre is [11+λ,21+λ] and radius = √(11+λ)2+(21+λ)2−1−λ1+λ=√4+λ21+λ. Since it touches the line x + 2y = 0, hence Radius = Perpendicular from centre to the line i.e., ∣∣∣11+λ+221+λ√12+22∣∣∣ = √4+λ21+λ ⇒√5=√4+λ2⇒=±1 λ=−1 cannot be possible in case of circle. So λ=1. Thus, from (i) x2+y2−x−2y=0 is the required equation of the circle.  Suggest Corrections  0      Similar questions  Explore more