Q90 / 180 · KCET 2011 · Mathematics
The centre of a circle which cuts $x^{2}+y^{2}+6 x-1=0, \quad x^{2}+y^{2}-3 y+2=0$ and $x^{2}+y^{2}+x+y-3=0$ orthogonally is
- A$\left(\frac{1}{7}, \frac{9}{7}\right)$
- B$\left(-\frac{1}{7},-\frac{9}{7}\right)$
- C$\left(\frac{1}{7},-\frac{9}{7}\right)$
- Correct answerD$\left(-\frac{1}{7}, \frac{9}{7}\right)$
Solution
The given circles
Let $\quad S_{1}=x^{2}+y^{2}+6 x-1=0$,
Centre $C_{1}=(-3,0)$, Radius $R_{1}=\sqrt{10}$
$$
S_{2} \equiv x^{2}+y^{2}-3 y+2=0,
$$
$$
\begin{gathered}
C_{2}=(0,3 / 2), R_{2}=1 / 2 \\
S_{3} \equiv x^{2}+y^{2}+x+y-3-0, \\
C_{3}=(-1 / 2,-1 / 2), R_{3}=\sqrt{7 / 2}
\end{gathered}
$$
Let $S=x^{2}+y^{2}+2 g x+2 f y+c=0$ be equation of circle which cut orthogonal all three circles. Then, by condition of orthogonality
$g+f=c-3$
On solving Eqs. (i), (ii) and (iii), we get
$$
g=1 / 7, f=-9 / 7
$$
So, centre is $(-g,-f)=(-1 / 7,9 / 7)$
Hence, the correct answer is option: 'D'.