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I was trying to solve this problem:
The point $A$ has coordinates $(5, 16)$ and the point $B$ has coordinates $(-4,4)$. The variable $P$ has coordinate $(x,y)$ and moves on a path such that $ AP = 2BP$. Show that the Cartesian equation of the path of the $P$ is:
$$(x+7)^2 +y^2 = 100\tag{*}$$ So, what I did is to found a point on the circular path, and shows that the relationship holds. However, the actual answer is to let: $$(x-5)^2 +(y-16)^2 = 4(x+4)^2 +4(y-4)^2$$ This also just means $AP=2BP$.
However, it will simplify to the original circle equation $(*)$. I literally don't know why. Even if $AP=2BP$ , why would this simplify to $(*)$. What's the mechanism behind this.

Thank you very much for you guys reply.

Alain Remillard
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Henry Cai
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  • Please help me guys, I am really confused by this, this is driving me crazy... – Henry Cai Jan 25 '20 at 20:39
  • , I finally got it. This is becuase A and 2B are constants. After simplification, we would get an equation for P, finally. Is this so? Because A is just equal to 2B? – Henry Cai Jan 25 '20 at 20:41

2 Answers2

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The locus of the points such that the ratio of their distance to two given points is constant is a circle:

$$\frac{\sqrt{(x-x_1)^2+(y-y_1)^2}}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=\lambda.$$

This is because

$$(x-x_1)^2+(y-y_1)^2-\lambda^2((x-x_0)^2+(y-y_0)^2)=0$$ is the equation of a conic, such that

  • the coefficients of $x^2,y^2$ are equal, and

  • with no cross term $xy$.

These are the conditions to have a circle.

When $\lambda=1$, the quadratic terms cancel each other, giving the equation of a line (the mediatrix).

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I'll suppose you use the usual euclidian distance between two points. Then the distance between $A$ and $P$ is $$d(A, P) =\sqrt{(x-5)^2+(y-16)^2}$$ Same goes for the distance between $B$ and $P$. $$d(B, P)=\sqrt{(x-(-4))^2+(y-4)^2}$$ Since $d(A, P)=D(B, P)$, we have $$\sqrt{(x-5)^2+(y-16)^2} =2\times\sqrt{(x+4)^2+(y-4)^2}$$ Squaring each side $$(x-5)^2 +(y-16)^2 = 4\left((x+4)^2 +(y-4)^2\right)$$ Developping each squares $$x^2-10x+25+y^2-32y+256 = 4x^2+32x+64+4y^2-32y+64$$ Bring all variables to the right, the rest to the left. $$153=3x^2+42x+3y^2$$ Since the coefficients of $x^2$ and $y^2$ are the same, it is a circle. Divide every term by $3$. $$51=x^2+14x+y^2$$ Complete the square for $x$ $$51+49=x^2+14x+49+y^2$$ $$100=(x+7)^2+y^2$$

Alain Remillard
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