Question

A circle $S$ cuts three circles
$x^2+y^2-4x-2y+4=0$
$x^2+y^2-2x-4y+1=0$
and $x^2+y^2+4x+2y+1=0$ 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
$S\equiv x^2+y^2+2gx+2fy+c=0$
Since, this circle cuts orthogonally to the circles
$x^2+y^2-4x-2y+4=0$
$x^2+y^2-2x-4y+1=0$
and $x^2+y^2+4x+2y+1=0$, respectively.
$\therefore 2g(2)+2f(1)=c+4...(i)$
$2g(1)+2f(2)=c+1...(ii)$
and $2g(-2)+2f(-1)=1+c...(iii)$
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 $S=\sqrt{{\left(\frac{3}{4}\right)}^2+{\left(\frac{-3}{4}\right)}^2+\frac{10}{4}}$
$=\sqrt{\frac{9}{16}+\frac{9}{16}+\frac{10}{4}}=\sqrt{\frac{58}{16}}=\sqrt{\frac{29}{8}}$