Question

Find 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 centre of the circle numerically less then or equal to $8$

Answer 1

Since $C(\alpha ,\beta )$ 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 $4x+3y-42=0$

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

Solving them, we get

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

$\therefore 6\beta =18$ or $\beta =3$ by taking the positive sign

and $8\alpha =66$ or $\alpha =\frac{66}{8}=\frac{33}{4}$ by taking negative sign.

Given $\left|\alpha \right|\le 8\Rightarrow -8\le \alpha \le 8$

$\therefore \alpha \ne \frac{33}{4}$

Put $\beta =3$ in equations $\left(1\right)$ and $\left(3\right)$ and equating, we get

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

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

$\Rightarrow 9{\alpha }^2+164\alpha -364=0$ which is quadratic in $\alpha$

$\therefore \alpha =\frac{-164\pm \sqrt{{164}^2-36\times -364}}{2\times 9}=\frac{-164\pm \sqrt{{164}^2+36\times 364}}{18}$

$=\frac{-164\pm \sqrt{40000}}{18}=\frac{-164\pm 200}{18}=2,\frac{-182}{9}$

But $-8\le \alpha \le 8$ $\therefore \alpha =2$

Now $r^2={(\alpha -2)}^2+{(3-8)}^2={(2-2)}^2+{(3-8)}^2=25$ where $r$ is the radius of the circle

Hence the required equation of the required circle is

${(x-2)}^2+{(y-3)}^2=25$

$x^2+y^2-4x-6y+4+9-25=0$ or $x^2+y^2-4x-6y-12=0$