Question

A curve is traced by a point P(x, y) which moves such that its distance from the point A(1, -2) is twice its distance from the point B(4, -3). Determine the equation of the curve.

Answer 1

`|PA|=√((1-x)^2+(-2-y)^2)\\\\|PB|=√((4-x)^2+(-3-y)^2)\\\\|PA|=2|PB|\\\\√((1-x)^2+(-2-y)^2)=2√((4-x)^2+(-3-y)^2)\\\\(1-x)^2+(-2-y)^2=4[(4-x)^2+(-3-y)^2]\\\\1-2x+x^2+4+4y+y^2=4(16-8x+x^2+9+6y+y^2)`


`x^2+y^2-2x+4y+5=4(x^2+y^2-8x+6y+25)\\\\x^2+y^2-2x+4y+5=4x^2+4y^2-32x+24y+100\\\\x^2-4x^2-2x+32x+y^2-4y^2+4y-24y+5-100=0\\\\-3x^2+30x-3y^2-20y-95=0\ \ \ \ /:(-3)\\\\x^2-10x+y^2+(20)/(3)y+(95)/(3)=0\\\\x^2-2x\cdot5+5^2-5^2+y^2+2y\cdot(10)/(3)+\left((10)/(3)\right)^2-\left((10)/(3)\right)^2=-(95)/(3)`


`(x-5)^2+(y+(10)/(3))^2=-(95)/(3)+(100)/(9)+25\\\\(x-5)^2+(y+(10)/(3))^2=-(285)/(9)+(100)/(9)+(225)/(9)\\\\(x-5)^2+(y+(10)/(3))^2=(40)/(9)\\\\it's\ the\ circle\\\\center:(5;-(10)/(3))\\\\radius:r=\sqrt(40)/(9)=(√(4\cdot10))/(\sqrt9)=(2√(10))/(3)`