Question

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

Answer 1

Let the centre of the circle be $(a,b)$
${(a-b)}^2+{(b-8)}^2=r^2..........\left(i\right)$
$\left|\frac{4a-3b-24}{5}\right|=\left|\frac{4a+3b-42}{5}\right|=r......\left(ii\right)$
$\Rightarrow$ $4a-3b-24=\pm (4a+3b-42).......\left(iii\right)$
$6b=18$ $\Rightarrow$ $b=3$
from $\left(i\right)$ ${(a-2)}^2=r^2-25$
from $\left(ii\right)$ $4a-3b-24=5r$
$4a=5r+33$
${(a-2)}^2={\left(\frac{4a-33}{5}\right)}^2-25$
${(a-2)}^2=(\frac{4a-33}{5}-5)(\frac{4a-33}{5}+5)$
${(a-2)}^2=\left(\frac{4a-58}{5}\right)\left(\frac{4a-8}{5}\right)$
$25{(a-2)}^2=4.2(2a-29)(a-2)$
$25(a-2)=8(2a-29)$ or $a-2=0$ $\Rightarrow$ $a=2$
Also $9a=-182$
$a=\frac{=182}{9}$
$\therefore$ $a=2,b=3,r=5$
and $a=\frac{-182}{9},b=3,r=\frac{205}{9}$
(from -ve sign in $\left(iii\right)$ $8a=66$
$\Rightarrow$ $a=33/4>8$