Find the equation of a locus...(Read More)

Question

Find the equation of the locus of a point which moves so that it's distance from (4,-3) is always one-half its distance from (-1,-1).

Answer 1

The writing is almost automatic: $$\sqrt{(x-4)^2+(y+3)^2}=\frac{1}{2}\sqrt{(x+1)^2+(y+1)^2}.$$ You will find it worthwhile to simplify. Square both sides. After some manipulation, you will find that the locus is a circle, of which you can find the centre and radius.

Remark: If what I called almost automatic is not quite as automatic as I think, please leave a message and I can amplify.

Answer 2

suppose A(4,-3) and B(-1,-1) and P(x,y) is your point. then 2PA=PB (PA and PB are euclidean distances).try it or to do easily try this 4$PA^2$ = $PB^2$ solve this equation it gives locus of P(x,y). $$\sqrt{(x-4)^2+(y+3)^2}=\frac{1}{2}\sqrt{(x+1)^2+(y+1)^2}.$$ $$\ 4{((x-4)^2+(y+3)^2)}=\ {(x+1)^2+(y+1)^2}.$$ $$\ 4{((x^2-8x+16)+(y^2+6y+9))}=\ {(x^2+2x+1)+(y^2+2y+1)}.$$ $$\ {(4x^2-32x+64)+(4y^2+24y+36)}=\ {x^2+y^2+2y+2x+2}.$$ $$\ {4x^2+4y^2+24y-32x+100}=\ {x^2+y^2+2y+2x+2}.$$ $$\ {3x^2+3y^2-34x+22y+98=0}.$$ this is a equation of a circle