Question

A straight line $AB$ is divided at $C$ so that $AC=3CB$. Circles are described on $AC$ and $CB$ as diameters and a common tangent meets $AB$ produced at $B$. Show that $BD$ is equal to the radius of the smaller circle.

Answer 1

We take the co-ordinates of $A$ and $B$ as $(-a,0)$ and $(a,0)$ respectively so that the mid-point of $AB$ is the origin and $AB$ the x-axis. If $(h,0)$ are the co-ordinates of $C$, then since $AC=3CB$, we have
$h=\frac{3.a+1.(-a)}{3+1}=\frac{1}{2}a$.
On $AC$ and $CB$ as diameters we draw two circles so that they touch each other at $C(\frac{1}{2}a,0)$. Their centres are $P(-\frac{1}{4}a,0)$ and $Q(\frac{3}{4}a,0)$ and radii are $\frac{3}{4}a$ and $\frac{1}{4}a$ respectively. Let a common tangent $RS$ meet $AB$ produced in $D$. Let $(\alpha ,0)$ be the co-ordinates of $D$. Then since $D$ divides $PQ$ externally in the ratio of the radii $\frac{3}{4}a:\frac{1}{4}a$
i.e. $3:1$, we have
$\alpha =\frac{3.\left(\frac{3}{4}a\right)-1.(-\frac{1}{4}a)}{3-1}=\frac{5}{4}a$.
$\therefore OD=\frac{5}{4}a$ and $OB=a$ so that
$BD=\frac{5}{\$}a-a=\frac{1}{4}a$.
Hence $BD$ is equal to the radius of the smaller circle.