Question

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

• A. $x^2+y^2+4x-6y-12=0$
• B. $x^2+y^2-4x+6y-12=0$
• C. $x^2+y^2-4x-6y-12=0$
• D. None of these

Answer 1

Answer: (C)

$x^2+y^2-4x-6y-12=0$

Let $C(\alpha ,\beta )$ be the centre of the circle.

Since the circle passes through the point $(2,8),$

$\therefore$ Radius of the circle $=\sqrt{{(\alpha -2)}^2+{(\beta -8)}^2}$

Since the circle touches the lines $4x-3y-24=0$ and $4x-3y-42=0,$

$\therefore \left|\frac{4\alpha -3\beta -24}{5}\right|=\left|\frac{4\alpha +3\beta -42}{5}\right|=\sqrt{{(\alpha -2)}^2+{(\beta -8)}^2}$ ...(1)

Solving them

$4\alpha -3\beta -24=\pm (4\alpha +3\beta -42)$ ...(3)

$\therefore 6\beta =18\Rightarrow \beta =3$ (taking positive sign)

and $8\alpha =66\Rightarrow \alpha =\frac{33}{4}$ (taking negative sign).

Given, $\left|\alpha \right|\le 8$ $\Rightarrow -8\le \alpha \le 8,\therefore \alpha \ne \frac{33}{4}$

Putting $\beta =3$ in equations (1) and (3) and equating, we get

${(4\alpha -33)}^2=[{(\alpha -2)}^2+25]$

$\Rightarrow 16{\alpha }^2-264\alpha +1089=25{\alpha }^2+725-100\alpha$

$\Rightarrow 9{\alpha }^2+164\alpha -364=0$

$\therefore \alpha =\frac{-164\pm \sqrt{{\left(164\right)}^2+36\times 364}}{18}=\frac{-164\pm 200}{18}=2,\frac{-182}{9}.$

But $-8\le \alpha \le 8,\therefore a=2.$

Now, ${\left(radius\right)}^2={(\alpha -2)}^2+{(3-8)}^2={(2-2)}^2+{(3-8)}^2=25$

Hence, the equation of the required circle is

${(x-2)}^2+{(y-3)}^2=25$$\Rightarrow x^2+y^2-4x-6y-12=0.$