The distance of P(x,y) to (3,0) is twice the distance of P from (0,3). Find the equation connecting x and y?

Question

please help. this question has been bugging me.

Answer 1

Using the distance formula:

2*sqrt[(x-3)^2+(y-0)^2]=sqrt[(x-0)^2 + (y-3)^2]

Multiplying out and simplifying gives:

x^2+y^2-8x+2y-9=0 (if I didn't make a calculation mistake somewhere along the way which is entirely possible :) )

Answer 2

Let u be the distance from point P to (3,0), and v be the distance from point P to (0,3).

Translating the word problem into an equation, we have

u = 2v

We have u = √[(x - 3)^2 + y^2] and v = √[x^2 + (y - 3)^2] using the distance formula. Substituting this, we now have the mathematical model

√[(x - 3)^2 + y^2] = 2√[x^2 + (y - 3)^2]

Squaring both sides and using algebraic identities:

(x - 3)^2 + y^2 = 4[x^2 + (y - 3)^2]

0 = 4[x^2 + (y - 3)^2] - [(x - 3)^2 + y^2]

0 = 4[x^2 + (y - 3)^2] - (x - 3)^2 - y^2

0 = 4x^2 + 4(y - 3)^2 - (x^2 - 6x + 9) - y^2

0 = 4x^2 + 4(y^2 - 6y + 9) - x^2 + 6x - 9 - y^2

0 = 3x^2 + 6x - 9 + 4y^2 - 24y + 36 - y^2

0 = 3x^2 + 6x + 27 + 3y^2 - 24y

0 = 3x^2 + 6x + 3y^2 - 24y + 27

This is an equation in the 2nd degree. We can complete the squares to simplify the format of the expression.

0 = 3(x^2 + 2x) + 3(y^2 - 8y) + 27

0 = 3(x^2 + 2x + 1) + 3(y^2 - 8y + 16) + 27 - 3 - 48

0 = 3(x + 1)^2 + 3(y - 4)^2 - 24

3(x + 1)^2 + 3(y - 4)^2 - 24 = 0

3(x + 1)^2 + 3(y - 4)^2 = 24

(x + 1)^2 + (y - 4)^2 = 8

And here's your equation that connects x and y. An equation of a circle.