Question

The equation of a circle passing through the point $(2,8)$, touching the lines
$4x-3y-24=0$ and $4x+3y-42=0$ and
having the $x$ coordinate of its centre less than or equal to $8$ is

• A. $x^2+y^2+2x-8y-8=0$
• B. $x^2+y^2-4x-6y-12=0$
• C. $x^2+y^2+4x-10y+4=0$
• D. $x^2+y^2-6x-4y-24=0$

Answer 1

Answer: (B)

By the property of distance,
$\left|\frac{4h-3k-24}{5}\right|=\left|\frac{4h+3,k-42}{5}\right|$
$=\sqrt{(h-2{)}^2+(k-8{)}^2}$
$\Rightarrow 4h-3k-24=\pm (4h+3k-42)$
either $4,h-3k-24=4h+3k-42$
$\Rightarrow 6k=18\Rightarrow k=3$
or $4h-3k-24=-4h-3k+42$
$\Rightarrow 8h=66$
$\Rightarrow h=\frac{66}{8}=\frac{33}{4}>8$
Now, $(4h-3k-24{)}^2=25\left[(h-2{)}^2+(k-8{)}^2\right]$
By solving this, we get, $h=2$
$\therefore$ Centre is $(2,3)$
Now, required circle is $(x-2{)}^2+(y-3{)}^2=25$
$\Rightarrow x^2+4-4x+y^2+9-6y=25$
$\Rightarrow x^2+y^2-4x-6x=12$