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