The circle \(S=0\) cuts the circle \(x^2+y^2-4 x+2 y-7=0\) orthogonally. If \((2,3)\) is the centre of the…
The circle \(S=0\) cuts the circle \(x^2+y^2-4 x+2 y-7=0\) orthogonally. If \((2,3)\) is the centre of the circle \(S=0\), then its radius is
2
1
3
4
Solution
Given that, \(S=0\) circle cuts the circle \(x^2+y^2-4 x+2 y-7=0\) orthogonally and centre of circle \(S=0\) is \((2,3)\).
As we know that, if two circles intersect orthogonally,
Then, \(2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime}\)
Here, \((g, f)=(2,3)\) and \(\left(g^{\prime}, f^{\prime}\right)=(2,-1)\)
\(\begin{array}{rlrl}
& c^{\prime} =-7 \\
& 2(-2)(2)+2(3)(-1)=c-7 \\
& \Rightarrow 8-6 =c-7 \\
& \Rightarrow 2 =c-7 \\
& \Rightarrow c =9 \\
& \text {Now, required radius } =\sqrt{(2)^2+(3)^2-9} \\
& =\sqrt{4+9-9}=\sqrt{4}=2
\end{array}\)