Question

The centre of the circle which cuts the three circles $x^2+y^2=9,x^2+y^2-2x+4=0$ and $x^2+y^2-4y+5=0$ orthogonally is

• A. $(-13,-7)$
• B. $(-\frac{13}{2},-\frac{7}{2})$
• C. $(\frac{13}{2},\frac{7}{2})$
• D. $(13,7)$

Answer 1

The correct option is C $(\frac{13}{2},\frac{7}{2})$

As per the property,
The locus of centre of a circle cutting two given circles orthogonally
is the radical axis of two circles.
$\therefore s_1=x^2+y^2=9$
$s_2=x^2+y^2-2x+4=0$
$s_3=x^2+y^2-4y+5=0$
$\therefore$ one radical axis is
$s_1-s_2=0$
$2x=13--\left(1\right)$
& other radical axis is
$s_1-s_3=0$
$4y=14--\left(2\right)$
So, from intersection of (1) & (2)
$x=\frac{13}{2}y=\frac{14}{4}$
$x=\frac{13}{2},y=\frac{7}{2}$