Intersection point of 2x−3y+4=0 and x−2y+3=0 is (1,2)
Let image of A(2,3) is B(α,β) Since, P is the fixed point for given family of lines So, PB=PA (α−1)2+(β−2)2=(2−1)2+(3−2)2 (α−1)2+(β−2)2=1+1=2 (x−1)2+(y−2)2=(√2)2 Compare with (x−a)2+(y−b)2=r2 Therefore, given locus is a circle with centre (1,2) and radius √2.