Question

Find locus of a point which moves so that

it's distance from the point (2,1) is double its distance from (1,2)
​

Answer 1

hi,

Explanation:

Let say P=(x,y) a point of the locus

`Distance\ from\ P\ to\ (2,1)= √((x-2)^2+(y-1)^2) \\Distance\ from\ P\ to\ (1,2)= √((x-1)^2+(y-2)^2) \\\\√((x-2)^2+(y-1)^2) =2*√((x-1)^2+(y-2)^2) \\\\(x-2)^2+(y-1)^2=4*((x-1)^2+(y-2)^2)\\\\3x^2-4x+3y^2-14y+15=0\\\\`

`3x^2-4x+3y^2-14y+15=0\\3(x^2-2*(2)/(3) x)+3(y^2-2*(7)/(3)*y) +15=0\\3(x^2-2*(2)/(3)*x+(4)/(9))+3(y^2-2*(7)/(3)*y+(49)/(9) ) +15-(4)/(3)-(49)/(3)=0\\\\3(x-(2)/(3))^2+3(y-(7)/(3))^2-(8)/(3)=0\\\\\\\boxed{(x-(2)/(3))^2+(y-(7)/(3))^2=(8)/(9)}\\\\`

Locus is the circle of center (2/3,7/3) and radius =2√2 /3.