Let A be the point (3, 0) and circles with variable diameter AB touch the circle x^2 + y^2…

Question

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

Answer: 18

Accepted Answer

Let the variable point be $B(h,k)$.

We are told that the circle having $AB$ as diameter touches the fixed circle ${x}^{2}+{y}^{2}=36$ internally.

If a circle has endpoints of a diameter at $A(3,0)$ and $B(h,k)$, then its equation is

$(x-h)(x-3)+(y-k)(y-0)=0$

Expanding this, we get

${x}^{2}+{y}^{2}-(h+3)x-ky+3h=0$

So, the centre of this circle is

Centre $(\frac{h+3}{2},\frac{k}{2})$

Now, the fixed circle ${x}^{2}+{y}^{2}=36$ has centre ${C}_{1}(0,0)$ and radius $R=6$.

The variable circle (with diameter $AB$) has radius equal to half of $AB$, so

$r=\frac{1}{2}(\sqrt{(h-3{)}^{2}+{k}^{2}})$

Since the two circles touch internally, the distance between their centres equals the difference of radii:

$	herefore {C}_{1}{C}_{2}=|R-r|$

Here, ${C}_{2}(\frac{h+3}{2},\frac{k}{2})$. So,

$\sqrt{{(\frac{h+3}{2})}^{2}+{(\frac{k}{2})}^{2}}=|6-\frac{1}{2}(\sqrt{(h-3{)}^{2}+{k}^{2}})|$

On simplifying and writing the locus in terms of a general point $(x,y)$ (i.e., replacing $(h,k)$ by $(x,y)$), we get

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

This is the standard form of an ellipse: sum of distances from two fixed points (foci) is constant $=2a$.

So,

$2a=12\, 	herefore a=6$

Comparing, the foci are $(-3,0)$ and $(3,0)$.

The distance between the foci is $6$, so $2c=6$. Using $c=ae$, we write

$2ae=6$

Substituting $a=6$:

$12e=6$

So,

$e=\frac{1}{2}$

Finally,

$	herefore 72{e}^{2}=72(\frac{1}{4})=18$

Solutions · 1