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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

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B

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C

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D

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Solution

## The correct options are A B Equation family of circle is S+λL=0 x2+y2−2x+1+λ(x+2y−1)=0 x2+y2−x(x−λ)+2λy+(1−λ)=0 - - - - - -(1) centre of circle is (-g,-f) =(2−λ2,−λ) And radius of the circle is =√g2+f2−c=√(2−λ2)2+λ2−(1−λ) =12√4+λ2−4λ+4λ2−4+4λ =12√5λ2=√52|λ| Since, circle touches the line 2x−y+3=0, therefore perpendicular from the centre should be equal to radius. Centre (2−λ2,−λ) Perpendicular distance from a point to a line is ∣∣ ∣∣2.((2−λ)2)−(−λ)+3√4+1∣∣ ∣∣=√52|λ| |2.λ+λ+3|√5 ∣∣∣5√5∣∣∣=√52|λ| √5=√52|λ| |λ|=2 λ=±2 Substituting the value of λ in equation (1) When λ=2 x2+y2−x(2−2)+2(2)y+(1−2)=0 x2+y2+4y−1=0 When λ=−2 x2+y2−x(2+2)+2(−2)y+(1+2)=0 x2+y2−4x−4y+3=0 so,option A and B are correct.

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