Question

Find the equation of circles which touch the two lines $x-2y+4=0$ and $2x-y-8=0$ and also pass through the point (4 , -1).

Answer 1

Draw a diagram of the lines and plot the point (4 , -1). You will observe that centre lies on the bisector of obtuse angle whose equation can be found as usual to be
$x+y-12=0$

If (h,k) be the centre then $h+k--12=0$
Again $(h-4{)}^2+(k+1{)}^2=r^2={\left(\frac{h-2k+4}{\sqrt{4}}\right)}^2$

Put $K=12-h$
or $5[{(h-4)}^2+(13-h{)}^2]=(3h-20{)}^2$

$or{h}^2-50h+525=0$

or $(h-15)(h-35)=0$

$\therefore h=15,k=-3,orh=35,k=-23$

$\therefore centreare(15,-3)and{r}^2=125$

or $(35,-23)and{r}^2=1445$

Above gives the data of two circles.