Q71 / 180 · KCET 2009 · Mathematics
The length of the diameter of the circle which cuts three circles
$$
\begin{array}{r}
x^{2}+y^{2}-x-y-14=0 \\
x^{2}+y^{2}+3 x-5 y-10=0 \\
x^{2}+y^{2}-2 x+3 y-27=0
\end{array}
$$
orthogonally, is
- A8
- B6
- Correct answerC4
- D2
Solution
Let the equation of circle be
$$
x^{2}+y^{2}+2 g x+2 f y+c=0
$$
and centre $(-g,-f)$
Centres of given circles are
$$
\mathrm{C}_{1}\left(\frac{1}{2}, \frac{1}{2}\right), \mathrm{C}_{2}\left(-\frac{3}{2}, \frac{5}{2}\right), \mathrm{C}_{3}\left(1,-\frac{3}{2}\right)
$$
Since, the Eq. (A) cut the given circles orthogonally.
$\begin{array}{lll}\therefore & -g-f=c-14 & \ldots(\text { i) } \\ & 3 g-5 f=c-10 & \ldots(\text { (ii) } \\ \text { and } & -2 g+3 f=c-27 & \ldots \text { (iii) }\end{array}$
On solving Eqs. (i), (ii) and (iii), we get
$$
g=-3, \quad f=-4, \quad c=21
$$
$\therefore$ From Eq. (A), the circle is
$$
x^{2}+y^{2}-6 x-8 y+21=0
$$
$\therefore$ Length of diameter
$$
=2 \sqrt{(-3)^{2}+(-4)^{2}-21}=4
$$
Hence, the correct answer is option: 'C'.