Number of circles intersecting $x^2+y^2=4$, $x^2+y^2-2 x-3=0$ and $x^2+y^2-2 y-3=0$ orthogonally is
Number of circles intersecting $x^2+y^2=4$, $x^2+y^2-2 x-3=0$ and $x^2+y^2-2 y-3=0$ orthogonally is
$0$
$1$
$2$
$\infty$
Solution
$S_1 \equiv x^2+y^2=4$...(i)
$S_2 \equiv x^2+y^2-2 x-3=0$...(ii)
$S_3 \equiv x^2+y^2-2 y-3=0$...(iii)
Let $S_4 \equiv x^2+y^2+2 g x+2 f y+c$ be the circle which is orthogonal with the circles $S_1, S_2$ and $S_3$.
Condition of orthogonal,
$2\left(g g^{\prime}+f f^{\prime}\right)=c+c^{\prime}$
Now, for $S_1$ and $S_4$;
$g \times 0+f \times 0=c-4$
$\Rightarrow \quad c=4$
For $S_2$ and $S_4$
$2[g \times(-1)+f \times(0)]=4-3$
$\Rightarrow \quad g=-\frac{1}{2}$
For $S_3$ and $S_4$,
$2[g \times(0)+f \times(-1)]=4-3$
$\Rightarrow \quad f=-\frac{1}{2}$
Now, radius of $S_4$,
$r^2=g^2+f^2-c=\left(\frac{1}{4}+\frac{1}{4}-4\right)=-\frac{7}{2} < 0$
Thus, no such $S_4$ circle is possible which is orthogonal to $S_1, S_2, S_3$.