The correct option is B (12,32)
Given lines are
√3(y−1)=x−1⇒x−√3y+√3−1=0⋯(1)y−1=√3(x−1)⇒−√3x+y+√3−1=0⋯(2)
a1a2+b1b2=−√3−√3<0
So, the acute angle bisector is
x−√3y+√3−12=+(−√3x+y+√3−1)2⇒y=x
Let the centre of the circle be (α,α)
Point of intersection of (1) and (2), we get
−2y+2=0⇒y=1⇒x=1
Now, the equation of the circle is,
(x−α)2+(y−α)2=(1−α)2+(1−α)2⇒x2+y2−2αx−2αy+4α−2=0
Given circle is
x2+y2+4x−6y+5=0
So, the common chord is,
S1−S2=0⇒(4+2α)x+(2α−6)y+(7−4α)=0⇒(4x−6y+7)+2α(x+y−2)=0
This represent family of lines, so the fixed point is the point of intersection of
4x−6y+7=0 and x+y−2=0⇒y=32⇒x=12
Hence, the fixed point is
(12,32)