Let A be the point (3,0), and let circles with variable diameter AB touch the fixed circle…

Question

Let $A$ be the point $(3,0)$, and let circles with variable diameter $AB$ touch the fixed circle $x^2+y^2=36$ internally. Let $C$ be the locus of $B$. If the eccentricity of $C$ is $e$, then $72e^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 18

Accepted Answer

Let $B=(h,k)$. The circle on diameter $AB$ has centre $\left(\dfrac{h+3}{2},\dfrac{k}{2}\right)$ and radius $r=\dfrac12\sqrt{(h-3)^2+k^2}$.

The fixed circle $x^2+y^2=36$ has centre $O(0,0)$ and radius $R=6$. Internal tangency gives $OC_2 = R-r$:

$$\sqrt{\left(\frac{h+3}{2}\right)^2+\left(\frac k2\right)^2} = 6-\frac12\sqrt{(h-3)^2+k^2}$$

which simplifies (writing $(h,k)\to(x,y)$) to

$$\sqrt{(x+3)^2+y^2}+\sqrt{(x-3)^2+y^2}=12.$$

This is an ellipse with foci $(\pm3,0)$ and $2a=12\Rightarrow a=6$; also $2c=6\Rightarrow c=3$, so $e=\dfrac{c}{a}=\dfrac12$.

$$72e^2 = 72\times\frac14 = 18.$$

Solutions · 1