Question

The lines $x-2y+4=0$ and $2x-y-8=0$ are tangents to a circle. Find the equation of that circle if it passes through the point $(4,-1)$.

Answer 1

Let $(h,k)$ be the centre and $r$ the radius. The point $(h,k)$ lies on either of the two bisectors of the angle between tangents, i.e.
$x+y-12=0$ and $3x-3y-4=0$
$\therefore h+k-12=0.....\left(1\right)$
and $h-k=4/3.....\left(2\right)$
It passes through $(4,-1)$
$\therefore (h-4{)}^2+(k+1{)}^2=r^2$
Again $p=r$ gives
$\therefore {\left(\frac{h-2k+4}{\sqrt{5}}\right)}^2=(h-4{)}^2+(k+1{)}^2.....\left(3\right)$
Now put $k=12-h$ from $\left(1\right)$
$\therefore (h-24+2h+4{)}^2=5[(h-4{)}^2+(13-h{)}^2]$
$\therefore (3h-20{)}^2=5(2h^2-34h+177)$
or $h^2-50h+525=0$
$\therefore h=15,35$ and $k=-3,-23$
Hence the circles are
$(x-15{)}^2+(y+3{)}^2=125$
or $(x-35{)}^2+(y+23{)}^2=1445$
Again if you put $k=h-4/3$ from $\left(2\right)$ and proceed as above you will not find real values of $h$ and etc.