I'm struggling with this geometry question:

The fixed points A and B have coordinates $(-3a,0)$ and $(a,0)$ respectively. Find the equation of the locus of a point P which moves in the coordinate plane so that $AP = 3PB$. Show that the locus is a circle, S, which touches the axis of $y$ and has its centre at the point ($\frac{3}{2}a, 0$)

I managed to get this part. First I assumed it was a circle, and used the ratio to find $x$ in terms of $a$. Let the coordinates of P be $(x,y)$:

$$ \sqrt{(x+3a)^2+y^2}=3\sqrt{(a-x)^2+y^2} \\ (x+3a)^2 + y^2 = 9(a-x)^2+y^2\\ x = 3a $$

So if one of the coordinates of S is $(3a,0)$, and S also touches the y-axis, we know that its centre is $(\frac{3}{2}a,0)$. So the equation of S is:

$$ \left(x-\frac{3}{2}\right)^2+y^2 = \frac{9}{4} $$

Is that correct? My approach seems quite hokey. The next part of the question is this:

A point Q moves in such a way that the perpendicular distance of Q from the axis of y is equal to the length of a tangent from Q to the circle S. Find the equation of the locus of Q. Show that this locus is also the locus of points which are equidistant from the line $4x+3a = 0$ and the point $(\frac{3}{4},0)$.

I tried finding the distance between $x=0$ and $Q(x,y)$, and that gave a distance of $x$ - but how do I plug that into the equality:

$$ D = x = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} $$


I think you overworked here. The condition is

$$|AP|^2=9|BP|^2\iff (x+3a)^2+y^2=9\left[(x-a)^2+y^2\right]\iff$$

$$\iff 8x^2-24ax+8y^2=0\iff \left(x-\frac{3a}2\right)^2+y^2=\frac{9a^2}4$$

and we get that the points$\;P=(x,y)\;$ are the locus of circle with center $\;\left(\frac{3a}2\,,\,0\right)\;$ and radius $\;\frac{3a}2\;$ , and since radius = absolute value of $\;x$ - coordinate of circle, the circle is tangent to the $\;y$ - axis

Let $\;Q=(x,y)\;$ , then its distance from the $\;y$ - axis is $\;|x|\;$ , and the length of any of its two tangets to the above circle is


and thus we get the equation:


$$\iff -3ax+y^2=0\;\;(*)$$

Now, let us check what is the locus of all point whose distance to $\;4x+3a=0\iff x=-\frac{3a}4\;$ equals its distance to $\;\left(\frac{3a}4\,,\,0\right)\;$:

$$\left(x-\frac{3a}4\right)^2+y^2=\frac{(4x+3a)^2}{16}\iff x^2-\frac{3a}2x+\frac{9a^2}{16}+y^2=x^2+\frac{3a}2x+\frac{9a^2}{16}\iff$$

$$\iff -3ax+y^2=0\;\;(*)$$

and thus both conditions are identical

  • $\begingroup$ Ah thank you, that seems much simpler. How about the last section of the question with point Q? $\endgroup$
    – hohner
    Oct 23 '14 at 17:52
  • $\begingroup$ There must be a mistake, I believe: it must be the point $\;\left(\frac{3a}4\,,\,0\right)\;$ , not $\;\left(\frac34\,,\,0\right)\;$ ...check this, please. $\endgroup$
    – Timbuc
    Oct 23 '14 at 18:11
  • $\begingroup$ Why does $|AP|^2 = 9|BP|^2$ rather than $|AP|=9|BP|$? $\endgroup$
    – hohner
    Oct 24 '14 at 16:09
  • $\begingroup$ I squared the condition $\;|AP|=3|PB|\;$ $\endgroup$
    – Timbuc
    Oct 24 '14 at 23:14

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