Question

Find the equations of the circles passing through $(-4,3)$ and touching the lines $x+ y - 2$ and $x - y =2$.

Answer 1

Let the equation of the required circle be
$x^2 + y^2 + 2gx + 2fy +c = 0....$ (i) It passes through $(-4,3)$.
$\therefore 25-8 g+6 f+c=0 ....$(ii)
Since, circle touches the line $x+y-2=0$ and
$x-y-2=0 . $
$\therefore\left|\frac{-g-f-2}{\sqrt{2}}\right|=\left|\frac{-g+f-2}{\sqrt{2}}\right|=\sqrt{g^{2}+f^{2}-c} ....$ (iii)
Now, $\left|\frac{-g-f-2}{\sqrt{2}}\right|=\left|\frac{-g+f-2}{\sqrt{2}}\right|$
$\Rightarrow -g-f-2 =\pm(-g+f-2)$
$\Rightarrow -g-f-2 =-g+f-2 $
or $-g-f-2 =g-f+2$
$\Rightarrow f =0 $ or $ g=-2$
Case I When $f=0$
From Eq. (iii), we get
$\left|\frac{-g-2}{\sqrt{2}}\right|=\sqrt{g^{2}-c}$
$ \Rightarrow(g+2)^{2}=2\left(g^{2}-c\right) $
$\Rightarrow g^{2}-4 g-4-2 c=0 ....$(iv)
On putting $f=0$ in Eq. (ii). we get
$25-8 g+c=0 ....$ (v)
Eliminating $c$ between Eqs. (iv) and (v), we get
$ \Rightarrow g^{2}-20 g+46 =0 $
$ g =10 \pm 3 \sqrt{6} $ and $ c=55 \pm 24 \sqrt{6}$
On substituting the values of $g, f$ and $c$ in Eq. (i), we get
$x^{2}+y^{2}+2(10 \pm 3 \sqrt{6}) x+(55 \pm 24 \sqrt{6})=0$
Case II When $g=-2$
From Eq. (iii), we get
$\Rightarrow f^{2}=2\left(4+f^{2}-c\right) $
$\Rightarrow f^{2}-2 c+8=0 .....$(vi)
On putting $g=-2$ in Eq. (ii), we get
$c=-6 f-41$
On substituting $c$ in Eq. (vi), we get
$f^{2}+12 f+90=0$
This equation gives imaginary values of $f$.
Thus, there is no circle in this case.
Hence, the required equations of the circles are
$x^{2}+y^{2}+2(10 \pm 3 \sqrt{6}) x+(55 \pm 24 \sqrt{6})=0$