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Hint-The radical lines of three circles are concurrent in a point known as the radical centre.

Let us consider three circles${S_1},{S_2},{S_3}$, then radical centre

${S_1} - {S_2} = 0$

And, ${S_2} - {S_3} = 0$

Let us take a point $\left( {x,y} \right)$ on the radical centre such that,

${\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} - {\left( {x + 3} \right)^2} - {\left( {y + 2} \right)^2} = 36 - 49.....\left( 1 \right)$

And, ${\left( {x + 3} \right)^2} + {\left( {y + 2} \right)^2} - {\left( {x - 4} \right)^2} - {\left( {y + 5} \right)^2} = 49 - 64.....\left( 2 \right)$

If we solve $\left( 1 \right)$ and $\left( 2 \right)$ , we get,

Answer$ \Rightarrow $$x = \dfrac{{26}}{{25}}$, $y = \dfrac{{13}}{{50}}$

Note: Make sure that when you take the equations of ${S_1},{S_2},{S_3}$, you do not take the square of it but the equation as it is.

Let us consider three circles${S_1},{S_2},{S_3}$, then radical centre

${S_1} - {S_2} = 0$

And, ${S_2} - {S_3} = 0$

Let us take a point $\left( {x,y} \right)$ on the radical centre such that,

${\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} - {\left( {x + 3} \right)^2} - {\left( {y + 2} \right)^2} = 36 - 49.....\left( 1 \right)$

And, ${\left( {x + 3} \right)^2} + {\left( {y + 2} \right)^2} - {\left( {x - 4} \right)^2} - {\left( {y + 5} \right)^2} = 49 - 64.....\left( 2 \right)$

If we solve $\left( 1 \right)$ and $\left( 2 \right)$ , we get,

Answer$ \Rightarrow $$x = \dfrac{{26}}{{25}}$, $y = \dfrac{{13}}{{50}}$

Note: Make sure that when you take the equations of ${S_1},{S_2},{S_3}$, you do not take the square of it but the equation as it is.