Question

Find the radius of the smallest circle which touches the straight line $3x-y=6$ at (1,-3) and also touches the line $y=x$. compute upto one place of decimal only.

Answer 1

The two lines meet at B (3,3), slopes of AB and BC are 3 and 1.
$\therefore tan2\theta =\frac{3-1}{1+3}=\frac{1}{2}$
Also from the figure , $\therefore tan2\theta =\frac{r}{AB}=\frac{r}{2\sqrt{10}}\therefore A{B}^2=40$
Now $\frac{2tan\theta }{1-tan\theta }=\frac{1}{2}$
or $4tan\theta =1-ta{n}^2\theta$
Putting the value of $tan\theta$, we get
$\frac{2r}{\sqrt{10}}=1-\frac{r^2}{40}$
$\therefore r^2+8\sqrt{10}r-40=0$
$r=\frac{-8\sqrt{10}\pm \sqrt{640+160}}{2}$
$=-4\sqrt{10}\pm 10\sqrt{2}=10\sqrt{2}-4\sqrt{10}$
$=10\times 1.414-4\times 3.162=1.49=1.5$