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KCET · Maths · Circle

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

  1. A \(\left(\frac{1}{7}, \frac{9}{7}\right)\)
  2. B \(\left(-\frac{1}{7},-\frac{9}{7}\right)\)
  3. C \(\left(\frac{1}{7},-\frac{9}{7}\right)\)
  4. D \(\left(-\frac{1}{7}, \frac{9}{7}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left(-\frac{1}{7}, \frac{9}{7}\right)\)

Step-by-step Solution

Detailed explanation

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 \((-g,-f)=(-1 / 7,9 / 7)\)