Locus Question Help! (1 Viewer)

[SGSII]

Hey guys!

Could you please help me solve this locus question?

Show that the locus of a point, which moves so as always to be 3 times further from one fixed point than from another fixed point, is a circle.

Provide an explanation as wells.
Thanks

 

[superSAIyan2]

Well this is the only way i can think of doing it.
Let P(x,y) be the variable point and A(x1,y1) and B(x2,y2) bethe fixed points such that PA = 3PB
Consequently (PA)^2 = 9(PB)^2
Using distance formula
PA ^2 = (X-X1)^2 + (Y - Y1)^2 = x^2 -2xx1 + x1^2 + y^2 -2yy1 + y1^2
PB^2 = (X-X2)^2 + (Y-Y2)^2 = x^2 -2xx2 + x2^2 + y^2 -2yy2 + y2^2
apply this to the condition (PA)^2 = 9(PB)^2

and after simplyifying and bringing all terms to one side you get
8x^2 -2x(x1 + 9x2) + ( 9x2^2 - x1^2) + 8y^2 -2y(y1+ 9y1) + (9y1^2 - y2^2)

This is the locus of P and since coefficients of x^2 and y^2 are the same - it is a circle

 

[SGSII]

Well this is the only way i can think of doing it.
Let P(x,y) be the variable point and A(x1,y1) and B(x2,y2) bethe fixed points such that PA = 3PB
Consequently (PA)^2 = 9(PB)^2
Using distance formula
PA ^2 = (X-X1)^2 + (Y - Y1)^2 = x^2 -2xx1 + x1^2 + y^2 -2yy1 + y1^2
PB^2 = (X-X2)^2 + (Y-Y2)^2 = x^2 -2xx2 + x2^2 + y^2 -2yy2 + y2^2
apply this to the condition (PA)^2 = 9(PB)^2

and after simplyifying and bringing all terms to one side you get
8x^2 -2x(x1 + 9x2) + ( 9x2^2 - x1^2) + 8y^2 -2y(y1+ 9y1) + (9y1^2 - y2^2)

This is the locus of P and since coefficients of x^2 and y^2 are the same - it is a circle

Okie dokes, Thankyouu!

 

[superSAIyan2]

no worries, sorry if it look a bit confusing - with so many x and y and ^ signs everywhere. Lol

 

[planino]

sorry if it look a bit confusing

Apologise for this too, Cadbury.