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Let $S \equiv x^2+y^2-6 x-6 y+4=0$ and $S^{\prime} \equiv x^2+y^2-2 x-4 y+3=0$ be two circles. The centre of a circle of radius $\sqrt{14}$ and having same radical axes with either of $S=0$ or $S^{\prime}=0$ is
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Verified Answer
The correct answer is:
$\left(\frac{-19}{5}, \frac{-2}{5}\right)$
Let the equation of required circle is
$x^2+y^2+2 g x+2 f y+c=0$
and the circle $1, s=0$ and $s^1=0$ are of a coaxial system of circles, so
$\frac{g+3}{2}=\frac{f+3}{1}=\frac{c-4}{-1}=k$
(let)
$\begin{aligned} & \Rightarrow g=2 k-3, f=k-3 \text { and } c=4-k \\ & \text { so, the radius }=\sqrt{14} \text { (given) } \\ & \Rightarrow \quad \sqrt{g^2+f^2-c}=\sqrt{14} \Rightarrow g^2+f^2-c=14 \\ & \Rightarrow 4 k^2-12 k+9+k^2-6 k+9-4+k=14 \\ & \Rightarrow 5 k^2-17 k=0 \Rightarrow k=\frac{17}{5},(k \neq 0)\end{aligned}$
$\therefore$ Centre of the required circle is
$(-g,-f)=\left(3-\frac{34}{5}, 3-\frac{17}{5}\right)=\left(-\frac{19}{5}, \frac{-2}{5}\right)$
$x^2+y^2+2 g x+2 f y+c=0$
and the circle $1, s=0$ and $s^1=0$ are of a coaxial system of circles, so
$\frac{g+3}{2}=\frac{f+3}{1}=\frac{c-4}{-1}=k$
(let)
$\begin{aligned} & \Rightarrow g=2 k-3, f=k-3 \text { and } c=4-k \\ & \text { so, the radius }=\sqrt{14} \text { (given) } \\ & \Rightarrow \quad \sqrt{g^2+f^2-c}=\sqrt{14} \Rightarrow g^2+f^2-c=14 \\ & \Rightarrow 4 k^2-12 k+9+k^2-6 k+9-4+k=14 \\ & \Rightarrow 5 k^2-17 k=0 \Rightarrow k=\frac{17}{5},(k \neq 0)\end{aligned}$
$\therefore$ Centre of the required circle is
$(-g,-f)=\left(3-\frac{34}{5}, 3-\frac{17}{5}\right)=\left(-\frac{19}{5}, \frac{-2}{5}\right)$
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