Question

A circle $S$ cuts three circles
$x^{2}+y^{2}-4 \,x-2 \,y+4=0$
$x^{2}+y^{2}-2 \,x-4\, y+1=0$
and $x^{2}+y^{2}+4 x+2 y+1=0$ orthogonally. Then, the radius of $S$ is

• A. $\sqrt{\frac{29}{8}}$
• B. $\sqrt{\frac{28}{11}}$
• C. $\sqrt{\frac{29}{7}}$
• D. $\sqrt{\frac{29}{5}}$

Answer 1

Answer: (A)

Let the equation of circle be
$S \equiv x^{2}+y^{2}+2\, g x+2\, f y+c=0$
Since, this circle cuts orthogonally to the circles
$x^{2}+y^{2}-4 x-2\, y+4=0 $
$x^{2}+y^{2}-2 x-4\, y+1=0$
and $x^{2}+y^{2}+4 \,x+2\, y+1=0$, respectively.
$\therefore 2 \,g(2)+2 \,f(1) =c+4 \,\,\,...(i)$
$2\, g(1)+2\, f(2) =c+1\,\,\,...(ii)$
and $2 g(-2)+2 f(-1)=1+c\,\,\,...(iii)$
On solving Eqs. (i), (ii) and (iii), we get
$g=\frac{3}{4}, f=-\frac{3}{4}, c=\frac{-10}{4}$
$\therefore $ Equation of circle $S \equiv x^{2}+y^{2}+\frac{3}{2} x-\frac{3}{2} y-\frac{10}{4}=0$
$\therefore $ Radius of circle $S=\sqrt{\left(\frac{3}{4}\right)^{2}+\left(\frac{-3}{4}\right)^{2}+\frac{10}{4}}$
$=\sqrt{\frac{9}{16}+\frac{9}{16}+\frac{10}{4}}=\sqrt{\frac{58}{16}}=\sqrt{\frac{29}{8}}$