Geometry - Locus #1

The locus is a projection of a circle from a single point, it is therefore a circle.

If $\overline {AP}:\overline{PQ}=2:3$ then $\overline {AP}:\overline{AQ}=2:5$

Triangles $\triangle ABP$ and $\triangle ACQ$ are similar, with all dimentions, including the radius $\overline {CQ}$ , enlarged in the ration $2:5$ .

$\overline {CQ}=\dfrac52\overline{BP}=\dfrac52\times2=5$

The length $L$ of the larger circle is $L=2\pi\overline{CQ}=10\pi$

The answer is $\dfrac {60L}{\pi}=\dfrac{60\times10\pi}{\pi}=\boxed{600}$

Let $A(-5,0)$ , $B(0,0)$ and $C: x^2+y^2=1$ .

And since $P$ is on circle $C$ , we can say that $P(\sin\theta, \cos\theta)$ .

Since $Q$ is a point that divides $AP$ externally with a ratio of $5:3$ ,

$Q\left(\frac{5\sin\theta-3\times(-5)}{5-3},\frac{5\cos\theta-3\times0}{5-3}\right)$

$\therefore\quad Q\left(5\sin\theta+\frac{15}{2}, 5\cos\theta\right)$ .

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Given point satisfies $\left(x-\frac{15}{2}\right)^2+y^2=5^2$ , and since $\theta$ doesn't have a specified range,

The locus of $Q$ is a circle whose center is $\left(\frac{15}{2},0\right)$ and radius is $5$ .

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$\therefore\quad \frac{60L}{\pi}=\frac{60\cdot2\cdot5\cdot\pi}{\pi}=\boxed{600}$