Question

The equation of the circle which cuts the three circles $x^2+y^2=a^2,(x-g{)}^2+y^2=a^2$ and
$x^2+(y-f{)}^2=a^2$ orthogonally is

• A. $x^2+y^2-2gx-2fy+a^2=0$
• B. $x^2+y^2-gx-fy+a^2=0$
• C. $x^2+y^2-fx-gy+a^2=0$
• D. $x^2+y^2+gx+fy-a^2=0$

Answer 1

The correct option is B $x^2+y^2-gx-fy+a^2=0$

$s_1:x^2+y^2=a^2$
$s_2:(x-g{)}^2+y^2=a^2$
$s_3:x^2+(y-f{)}^2=a^2$
radical axis of $s_1$ and $s_2$ is $s_1-s_2=0$
$-2gx+g^2=0$
$x=\frac{g}{2}--\left(1\right)$
& radical axis of $s_1$ and $s_3$ is $s_3-s_1=0$
$-2fy+f^2=0$
$y=\frac{f}{2}--\left(2\right)$
So, centre of circle which is orthogonal to given
circle is radical centre $(\frac{g}{2},\frac{f}{2}).$
and radius of circle is $\sqrt{g\frac{2}{4}+\frac{f^2}{4}-a^2}$
So, $e{q}^n$ of circle is
$(x-\frac{g}{2}{)}^2+(y-\frac{f}{2}{)}^2=\frac{g^2}{4}+\frac{f^2}{4}-a^2$
$\Rightarrow x^2+y^2-gx-fy+a^2=0$