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
x2 + y2 − 4x − 4y + 3 = 0
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.