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
$x^{2}+y^{2}+2 g x+2 f y+c=0\,\,\,\,\,\,\,\,\,\,...(A)$
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.
$\therefore \,\,\,\,-g-f=c-14\,\,\,\,...(i)$
$3 g-5 f=c-10\,\,\,\,...(ii)$
and $\,\,\,\,-2 g+3 f=c-27\,\,\,\,...(iii)$
On solving Eqs. (i), (ii) and (iii), we get
$g=-3, f=-4, c=21$
$\therefore $ From Eq. (A), the circle is
$x^{2}+y^{2}-6 x-8 y+21=0$
$\therefore $ Length of diameter
$=2 \sqrt{(-3)^{2}+(-4)^{2}-21}=4$