Question

The length of the diameter of the circle which cuts three circles $x^2+y^2-x-y-14=0;$
$x^2+y^2+3x-5y-10=0;$
$x^2+y^2-2x+3y-27=0$
orthogonally, is

• A. 2
• B. 8
• C. 6
• D. 4

Answer 1

Answer: (D)

Let the equation of circle be
$x^2+y^2+2gx+2fy+c=0...(A)$
and centre $(-g,-f)$
Centres of given circles are
$C_1\left(\frac{1}{2},\frac{1}{2}\right),C_2\left(-\frac{3}{2},\frac{5}{2}\right),C_3\left(1,-\frac{3}{2}\right)$
Since, the Eq. $(A)$ cut the given circles orthogonally.
$\therefore -g-f=c-14...(i)$
$3g-5f=c-10...(ii)$
and $-2g+3f=c-27...(iii)$
On solving Eqs. (i), (ii) and (iii), we get
$g=-3,f=-4,c=21$
$\therefore$ From Eq. (A), the circle is
$x^2+y^2-6x-8y+21=0$
$\therefore$ Length of diameter
$=2\sqrt{(-3{)}^2+(-4{)}^2-21}=4$