The locus of a point equidistant from the intersection of lines $3x-7y+7=0$and $4x-6y+1=0$ is a
line parallel to $7x+13y+8=0$
bisector of the line $7x+13y+8=0$
You don’t need to crack your head, to solve the question.
The correct option is option B (Circle)
Now let me explain why
Where the two lines 3x –7y +7 = 0 and 4x –6y +1 = 0 intersect will be the centre of the circle which has radius r. The radius always equidistant to the circumference of the circle. I believe this explanation will drive home the point I am making.