Find the equation of a circle which cuts the circle $x^2+y^2-6 x+4 y-3=0$ orthogonally, while passing…
Find the equation of a circle which cuts the circle $x^2+y^2-6 x+4 y-3=0$ orthogonally, while passing through $(3,0)$ and touching the $Y$-axis.
$x^2+y^2+6 x+6 y+9=0$
$x^2+y^2-6 x-6 y+9=0$
$x^2+y^2-6 x+6 y-9=0$
$x^2+y^2+6 x-6 y-9=0$
Solution
When two circles intersects each other orthogonally, then
$2\left(g_1 g_2+f_1 f_2\right)=c_1 c_2$, where two circles are
$\begin{aligned}
& x^2+y^2+2 g_1 x+2 f_1 y+c_1=0 \text { and } \\
& x^2+y^2+2 g_2 x+2 f_2 y+c_2=0
\end{aligned}$
Let $c(h, k)$ be the centre of required circle which passes through $(3,0)$ and also touches $Y$-axis?
$\therefore$ Radius $=\sqrt{(h-3)^2+(k-0)^2}=|h|$
$\Rightarrow \quad(h-3)^2+k^2=h^2$
$\Rightarrow \quad k^2-6 h+9=0$ ...(i)
Required circle
$(x-h)^2+(y-k)^2=h^2$
$\Rightarrow \quad x^2+y^2-2 h x-2 k y+k^2=0$ ...(ii)
$\Rightarrow \quad g_1=-h_1, f_1=-k, c_1=k^2$
$\because$ Circle (ii) is intersected orthogonally by
$x^2+y^2-6 x+4 y-3=0$
$\Rightarrow \quad g_2=-3, f_2=2, c_2=-3$
$\therefore \quad 2\left(g_1 g_2+f_1 f_2\right)=c_1+c_2$
$\begin{array}{ll}\Rightarrow & 2[3 h-2 k]=-3+k^2 \\ \Rightarrow & 6 h-4 k=-3+k^2\end{array}$
$\Rightarrow \quad k^2-6 h-3=-4 k$
$\Rightarrow \quad-9-3=-4 k$ [from Eq. (i)]
$k=3$
$\begin{aligned} \therefore & & 3^2-6 h+9 & =0 \\ \Rightarrow & & h & =3\end{aligned}$
$\therefore \text { Center }=(3,3)$
and from Eq. (ii) required circle is
$x^2+y^2-6 x-6 y+9=0$