Question

A circle centered at $P$ with radius $4cm$ another circle centered at $Q$ with radius $16cm$ touch each other externally. A third circle with centre $R$ is drawn to touch the first two circles and one of thier common tangents as shown in the figure. Then, the radius of the circle with centre $R$ is

• A. $\frac{1}{3}cm$
• B. $2cm$
• C. $\frac{16}{9}cm$
• D. $\frac{4}{9}cm$

Answer 1

The correct option is C $\frac{16}{9}cm$

As shown in figure,

$PQ=4+16=20$

$PB=4$, $QC=20$

Refer the figure and constructions made in the figure.

Let $OB=x$ and $OC=y$

Let radius of smaller circle touching other two circles orthogonally is r.

$\therefore AB=RO=DC=r$

Now, $PA=PB-AB$

$\therefore PA=4-r$

Similarly, $PR=4+r$

In right angled triangle $△PAR$, by Pythagoras theorem,

${PR}^2={PA}^2+{AR}^2$

$\therefore {(4+r)}^2={(4-r)}^2+{AR}^2$

$\therefore 16+8r+r^2=16-8r+r^2+{AR}^2$

$\therefore {AR}^2=16r$

Taking square roots on both sides, we get,

$\therefore AR=4\sqrt{r}$

But, $AR=BO=x$

$\therefore x=4\sqrt{r}$

Now, $QD=QC-DC$

$\therefore QD=16-r$

Similarly, $QR=16+r$

In right angled triangle $△QDR$, by Pythagoras theorem, we get,

${QR}^2={QD}^2+{RD}^2$

$\therefore {(16+r)}^2={(16-r)}^2+{RD}^2$

$\therefore 256+32r+r^2=256-32r+r^2+{RD}^2$

$\therefore {RD}^2=64r$

Taking square roots n both sides,

$\therefore RD=8\sqrt{r}$

But, $RD=OC=y$

$\therefore y=8\sqrt{r}$

Now, consider right angled triangle PQS

By Pythagoras theorem,

${PQ}^2={PS}^2+{QS}^2$

$\therefore {\left(20\right)}^2={(x+y)}^2+{(16-4)}^2$

$\therefore 400={(x+y)}^2+144$

$\therefore {(x+y)}^2=400-144$

$\therefore {(x+y)}^2=256$

Taking square roots on both sides, we get,

$\therefore x+y=16$

$\therefore 4\sqrt{r}+8\sqrt{r}=16$

$\therefore 12\sqrt{r}=16$

$\therefore \sqrt{r}=\frac{16}{12}=\frac{4}{3}$

Squaring both sides, we get,

$\therefore r=\frac{16}{9}$

Thus, answer is option (C)