Question

Find the equation of circles passing through the point $(2,8)$, touching the lines $4x-3y-24=0$ and $4x+3y-42=0$ and having x-co-ordinate of the centre of the circle less than or equal to $8$.

Answer 1

The bisectors of the given tangents are $y=3$ and $x=66/8$ on which lie the centre of the circle. Since $x\le 8$,
$\therefore x=\frac{66}{8}$ is rejected. Hence let the centre be $(h,3)$ and radius be taken as $a$. The circle passes through $(2,8)$.
$\therefore (h-2{)}^2+(3-8{)}^2=a^2$
or $(h-2{)}^2+25=a^2.....(1)$
The given condition of tangency $p=r$ for any tangent gives
$4h-33=\pm 5a.....\left(2\right)$
Eliminating $a$ between $\left(1\right)$ and $\left(2\right)$, we get
$9h^2+164h-364=0$
or $(h-2)(9h+182)=0$.
$\therefore h=2$ and from $\left(2\right)a=5$.
$\therefore (x-2{)}^2+(y-3{)}^2=25$
or $x^2+y^2-4x-6y-12=0$.