Let BC be a fixed line segment in the plane . The locus of a point A such that the triangle ABC is isosceles , is (with finitely many possible exceptional points )

Options

A. a line
B. a circle
C. the union of a circle and a line
D. the union of two circles and a line

Answer

D

Explanation

Case (i) :

If `angleB = angleC`
locus of A is `bot` bisector of BC
So it is straight line
Case (ii) :

If `angleA = angleC`
BC fixed B (a , 0 ) , C(0, a)
BC = AB
So , `(x-a)^(2) + y^(2) = 2a^(2)`
Circle
Case (iii) :
`angleA = angleB`
AC = BC
`sqrt(h^(2) + (k-a)^(2)) = sqrt(2a^(2))`
`x^(2) + (y-a)^(2) = 2a^(2)`
also a circle
So union of two circle and a line .